Chapter 06 Permutations And Combinations
Every body of discovery is mathematical in form because there is no other guidance we can have  DARWIN
6.1 Introduction
Suppose you have a suitcase with a number lock. The number lock has 4 wheels each labelled with 10 digits from 0 to 9 . The lock can be opened if 4 specific digits are arranged in a particular sequence with no repetition. Some how, you have forgotten this specific sequence of digits. You remember only the first digit which is 7. In order to open the lock, how many sequences of 3digits you may have to check with? To answer this question, you may, immediately, start listing all possible arrangements of 9 remaining digits taken 3 at a time. But, this method will be tedious, because the number of possible sequences may be large. Here, in this Chapter, we shall learn some basic counting techniques which will
enable us to answer this question without actually listing 3digit arrangements. In fact, these techniques will be useful in determining the number of different ways of arranging and selecting objects without actually listing them. As a first step, we shall examine a principle which is most fundamental to the learning of these techniques.
6.2 Fundamental Principle of Counting
Let us consider the following problem. Mohan has 3 pants and 2 shirts. How many different pairs of a pant and a shirt, can he dress up with? There are 3 ways in which a pant can be chosen, because there are 3 pants available. Similarly, a shirt can be chosen in 2 ways. For every choice of a pant, there are 2 choices of a shirt. Therefore, there are $3 \times 2=6$ pairs of a pant and a shirt.
Let us name the three pants as $P_1, P_2, P_3$ and the two shirts as $S_1, S_2$. Then, these six possibilities can be illustrated in the Fig. 6.1.
Fig 6.1
Let us consider another problem of the same type.
Sabnam has 2 school bags, 3 tiffin boxes and 2 water bottles. In how many ways can she carry these items (choosing one each).
A school bag can be chosen in 2 different ways. After a school bag is chosen, a tiffin box can be chosen in 3 different ways. Hence, there are $2 \times 3=6$ pairs of school bag and a tiffin box. For each of these pairs a water bottle can be chosen in 2 different ways.
Hence, there are $6 \times 2=12$ different ways in which, Sabnam can carry these items to school. If we name the 2 school bags as $B_1, B_2$, the three tiffin boxes as $T_1, T_2, T_3$ and the two water bottles as $W_1, W_2$, these possibilities can be illustrated in the Fig. 6.2.
Fig 6.2
In fact, the problems of the above types are solved by applying the following principle known as the fundamental principle of counting, or, simply, the multiplication principle, which states that
“If an event can occur in $m$ different ways, following which another event can occur in $n$ different ways, then the total number of occurrence of the events in the given order is $m \times n$.”
The above principle can be generalised for any finite number of events. For example, for 3 events, the principle is as follows:
‘If an event can occur in $m$ different ways, following which another event can occur in $n$ different ways, following which a third event can occur in $p$ different ways, then the total number of occurrence to “the events in the given order is $m \times n \times p$.”
In the first problem, the required number of ways of wearing a pant and a shirt was the number of different ways of the occurence of the following events in succession:
(i) the event of choosing a pant
(ii) the event of choosing a shirt.
In the second problem, the required number of ways was the number of different ways of the occurence of the following events in succession:
(i) the event of choosing a school bag
(ii) the event of choosing a tiffin box
(iii) the event of choosing a water bottle.
Here, in both the cases, the events in each problem could occur in various possible orders. But, we have to choose any one of the possible orders and count the number of different ways of the occurence of the events in this chosen order.
Example 1 Find the number of 4 letter words, with or without meaning, which can be formed out of the letters of the word ROSE, where the repetition of the letters is not allowed.
Solution There are as many words as there are ways of filling in 4 vacant places $\square \square \square \square$ by the 4 letters, keeping in mind that the repetition is not allowed. The first place can be filled in 4 different ways by anyone of the 4 letters R,O,S,E. Following which, the second place can be filled in by anyone of the remaining 3 letters in 3 different ways, following which the third place can be filled in 2 different ways; following which, the fourth place can be filled in 1 way. Thus, the number of ways in which the 4 places can be filled, by the multiplication principle, is $4 \times 3 \times 2 \times 1=24$. Hence, the required number of words is 24 .
Note  If the repetition of the letters was allowed, how many words can be formed? One can easily understand that each of the 4 vacant places can be filled in succession in 4 different ways. Hence, the required number of words $=4 \times 4 \times 4 \times 4=256$.
Example 2 Given 4 flags of different colours, how many different signals can be generated, if a signal requires the use of 2 flags one below the other?
Solution There will be as many signals as there are ways of filling in 2 vacant places $\begin{array}{l} \hline \quad \\ \hline \\ \hline \end{array}$ in succession by the 4 flags of different colours. The upper vacant place can be filled in 4 different ways by anyone of the 4 flags; following which, the lower vacant place can be filled in 3 different ways by anyone of the remaining 3 different flags. Hence, by the multiplication principle, the required number of signals $=4 \times 3=12$.
Example 3 How many 2 digit even numbers can be formed from the digits $1,2,3,4,5$ if the digits can be repeated?
Solution There will be as many ways as there are ways of filling 2 vacant places $\square \square$ in succession by the five given digits. Here, in this case, we start filling in unit’s place, because the options for this place are 2 and 4 only and this can be done in 2 ways; following which the ten’s place can be filled by any of the 5 digits in 5 different ways as the digits can be repeated. Therefore, by the multiplication principle, the required number of two digits even numbers is $2 \times 5$, i.e., 10 .
Example 4 Find the number of different signals that can be generated by arranging at least 2 flags in order (one below the other) on a vertical staff, if five different flags are available.
Solution A signal can consist of either 2 flags, 3 flags, 4 flags or 5 flags. Now, let us count the possible number of signals consisting of 2 flags, 3 flags, 4 flags and 5 flags separately and then add the respective numbers.
There will be as many 2 flag signals as there are ways of filling in 2 vacant places $\begin{array}{l} \hline \quad \\ \hline \\ \hline \end{array}$ in succession by the 5 flags available. By Multiplication rule, the number of ways is $5 \times 4=20$.
Similarly, there will be as many 3 flag signals as there are ways of filling in 3 vacant places $\begin{array}{l} \hline \quad \\ \hline \\ \hline \\ \hline \end{array}$ in succession by the 5 flags.
The number of ways is $5 \times 4 \times 3=60$.
Continuing the same way, we find that
The number of 4 flag signals $=5 \times 4 \times 3 \times 2=120$
and the number of 5 flag signals $=5 \times 4 \times 3 \times 2 \times 1=120$
Therefore, the required no of signals $=20+60+120+120=320$.
EXERCISE 6.1
1. How many 3digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
(i) repetition of the digits is allowed?
(ii) repetition of the digits is not allowed?
Show Answer
Answer :
(i) There will be as many ways as there are ways of filling 3 vacant places
$ \begin{array}{ccc} \hline & & \\ \hline \end{array} $
in succession by the given five digits. In this case, repetition of digits is allowed. Therefore, the units place can be filled in by any of the given five digits. Similarly, tens and hundreds digits can be filled in by any of the given five digits.
Thus, by the multiplication principle, the number of ways in which threedigit numbers can be formed from the given digits is $5 \times 5 \times 5=125$
(ii) In this case, repetition of digits is not allowed. Here, if units place is filled in first, then it can be filled by any of the given five digits. Therefore, the number of ways of filling the units place of the threedigit number is 5 .
Then, the tens place can be filled with any of the remaining four digits and the hundreds place can be filled with any of the remaining three digits.
Thus, by the multiplication principle, the number of ways in which threedigit numbers can be formed without repeating the given digits is $5 \times 4 \times 3=60$
2. How many 3digit even numbers can be formed from the digits 1,2,3,4, 5, 6 if the digits can be repeated?
Show Answer
Answer :
There will be as many ways as there are ways of filling 3 vacant places
$\square \square \square$ in succession by the given six digits. In this case, the units place can be filled by 2 or 4 or 6 only i.e., the units place can be filled in 3 ways. The tens place can be filled by any of the 6 digits in 6 different ways and also the hundreds place can be filled by any of the 6 digits in 6 different ways, as the digits can be repeated.
Therefore, by multiplication principle, the required number of three digit even numbers is $3 \times 6 \times 6=108$
3. How many 4letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
Show Answer
Answer :
There are as many codes as there are ways of filling 4 vacant places $\square \square \square \square$ in succession by the first 10 letters of the English alphabet, keeping in mind that the repetition of letters is not allowed.
The first place can be filled in 10 different ways by any of the first 10 letters of the English alphabet following which, the second place can be filled in by any of the remaining letters in 9 different ways. The third place can be filled in by any of the remaining 8 letters in 8 different ways and the fourth place can be filled in by any of the remaining 7 letters in 7 different ways.
Therefore, by multiplication principle, the required numbers of ways in which 4 vacant places can be filled is $10 \times 9 \times 8 \times 7=5040$
Hence, 5040 fourletter codes can be formed using the first 10 letters of the English alphabet, if no letter is repeated.
4. How many 5digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
Show Answer
Answer :
It is given that the 5digit telephone numbers always start with 67.
Therefore, there will be as many phone numbers as there are ways of filling 3 vacant places
$ \begin{array}{ccc} \hline 6 & 7\\ \hline \end{array} $
by the digits $09$, keeping in mind that the digits cannot be repeated.
The units place can be filled by any of the digits from 0  9, except digits 6 and 7 . Therefore, the units place can be filled in 8 different ways following which, the tens place can be filled in by any of the remaining 7 digits in 7 different ways, and the hundreds place can be filled in by any of the remaining 6 digits in 6 different ways.
Therefore, by multiplication principle, the required number of ways in which 5digit telephone numbers can be constructed is $8 \times 7 \times 6=336$
5. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
Show Answer
Answer :
When a coin is tossed once, the number of outcomes is 2 (Head and tail) i.e., in each throw, the number of ways of showing a different face is 2 .
Thus, by multiplication principle, the required number of possible outcomes is $2 \times 2 \times 2=8$
6. Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
Show Answer
Answer :
Each signal requires the use of 2 flags.
There will be as many flags as there are ways of filling in 2 vacant places $\square$ in succession by the given 5 flags of different colours.
The upper vacant place can be filled in 5 different ways by any one of the 5 flags following which, the lower vacant place can be filled in 4 different ways by any one of the remaining 4 different flags.
Thus, by multiplication principle, the number of different signals that can be generated is $5 \times 4=20$
6.3 Permutations
In Example 1 of the previous Section, we are actually counting the different possible arrangements of the letters such as ROSE, REOS, …, etc. Here, in this list, each arrangement is different from other. In other words, the order of writing the letters is important. Each arrangement is called a permutation of 4 different letters taken all at a time. Now, if we have to determine the number of 3letter words, with or without meaning, which can be formed out of the letters of the word NUMBER, where the repetition of the letters is not allowed, we need to count the arrangements NUM, NMU, MUN, NUB, …, etc. Here, we are counting the permutations of 6 different letters taken 3 at a time. The required number of words $=6 \times 5 \times 4=120$ (by using multiplication principle).
If the repetition of the letters was allowed, the required number of words would be $6 \times 6 \times 6=216$.
Definition 1 A permutation is an arrangement in a definite order of a number of objects taken some or all at a time.
In the following subsection, we shall obtain the formula needed to answer these questions immediately.
6.3.1 Permutations when all the objects are distinct
Theorem 1 The number of permutations of $n$ different objects taken $r$ at a time, where $0<r \leq n$ and the objects do not repeat is $n(n1)(n2) \ldots(nr+1)$, which is denoted by ${ }^{n} P_r$.
Proof There will be as many permutations as there are ways of filling in $r$ vacant places $ \underset{\leftarrow r \text{ vacant places} \rightarrow}{\Large{\square \square \square \cdots }} \Large{\square}$ by
the $n$ objects. The first place can be filled in $n$ ways; following which, the second place can be filled in $(n1)$ ways, following which the third place can be filled in $(n2)$ ways,…, the $r$ th place can be filled in $(n(r1))$ ways. Therefore, the number of ways of filling in $r$ vacant places in succession is $n(n1)(n2) \ldots(n(r1))$ or $n(n1)(n2) \ldots(nr+1)$
This expression for ${ }^{n} P$ is cumbersome and we need a notation which will help to reduce the size of this expression. The symbol $n$! (read as factorial $n$ or $n$ factorial ) comes to our rescue. In the following text we will learn what actually $n$! means.
6.3.2 Factorial notation
The notation $n$! represents the product of first $n$ natural numbers, i.e., the product $1 \times 2 \times 3 \times \ldots \times(n1) \times n$ is denoted as $n$ !. We read this symbol as ’ $n$ factorial’. Thus, $1 \times 2 \times 3 \times 4 \ldots \times(n1) \times n=n$ !
$ \begin{aligned} & 1=1 ! \\ & 1 \times 2=2 ! \\ & 1 \times 2 \times 3=3 ! \\ & 1 \times 2 \times 3 \times 4=4 \text{ ! and so on. } \end{aligned} $
We define $0 !=1$
We can write $5 !=5 \times 4 !=5 \times 4 \times 3 !=5 \times 4 \times 3 \times 2$ !
$$ =5 \times 4 \times 3 \times 2 \times 1 \text{ ! } $$
Clearly, for a natural number $n$
$$ \begin{array}{rlrl} n ! & =n(n1) ! & \\ & =n(n1)(n2) ! & & \text { [ provided } n \geq 2] \\ & =n(n1)(n2)(n3) ! & & \text { [ provided } n \geq 3] \end{array} $$
and so on.
Example 5 Evaluate
(i) 5 !
(ii) 7 !
(iii) $7 !5$ !
Solution
(i) $5 !=1 \times 2 \times 3 \times 4 \times 5=120$
(ii) 7 ! $=1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7=5040$
and
(iii) $7 !5 !=5040120=4920$.
Example 6 Compute (i) $\frac{7 !}{5 !}$
(ii) $\frac{12 !}{(10 !)(2 !)}$
Solution
(i) We have $\frac{7 !}{5 !}=\frac{7 \times 6 \times 5 !}{5 !}=7 \times 6=42$
and
(ii) $\frac{12 !}{(10 !)(2 !)}=\frac{12 \times 11 \times(10 !)}{(10 !) \times(2)}=6 \times 11=66$.
Example 7 Evaluate $\frac{n !}{r !(nr) !}$, when $n=5, r=2$.
Solution We have to evaluate $\frac{5 !}{2 !(52) !}($ since $n=5, r=2)$
We have $\quad \frac{5 !}{2 !(52) !}=\frac{5 !}{2 ! \times 3 !}=\frac{5 \times 4}{2}=10$
Example 8 If $\frac{1}{8 !}+\frac{1}{9 !}=\frac{x}{10 !}$, find $x$.
Solution We have $\frac{1}{8 !}+\frac{1}{9 \times 8 !}=\frac{x}{10 \times 9 \times 8 !}$
Therefore $1+\frac{1}{9}=\frac{x}{10 \times 9}$ or $\frac{10}{9}=\frac{x}{10 \times 9}$
So
$ x=100 . $
EXERCISE 6.2
1. Evaluate
(i) 8 !
(ii) 4 ! 3 !
Show Answer
Answer :
(i) $8 !=1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8=40320$
(ii) 4 ! $=1 \times 2 \times 3 \times 4=24$
$3 !=1 \times 2 \times 3=6$
$\therefore 4$ !  3 ! $=246=18$
2. Is $3 !+4 !=7 !$ ?
Show Answer
Answer :
$3 !=1 \times 2 \times 3=6$
$4 !=1 \times 2 \times 3 \times 4=24$
$\therefore 3 !+4 !=6+24=30$
$7 !=1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7=5040$
$\therefore 3 !+4 ! \neq 7$ !
3. Compute $\dfrac{8 !}{6 ! \times 2 !}$
Show Answer
Answer :
$\dfrac{8 !}{6 ! \times 2 !}=\dfrac{8 \times 7 \times 6 !}{6 ! \times 2 \times 1}=\dfrac{8 \times 7}{2}=28$
4. If $\dfrac{1}{6 !}+\dfrac{1}{7 !}=\dfrac{x}{8 !}$, find $x$
Show Answer
Answer :
$\dfrac{1}{6 !}+\dfrac{1}{7 !}=\dfrac{x}{8 !}$
$\Rightarrow \dfrac{1}{6 !}+\dfrac{1}{7 \times 6 !}=\dfrac{x}{8 \times 7 \times 6 !}$
$\Rightarrow \dfrac{1}{6 !}(1+\dfrac{1}{7})=\dfrac{x}{8 \times 7 \times 6 !}$
$\Rightarrow 1+\dfrac{1}{7}=\dfrac{x}{8 \times 7}$
$\Rightarrow \dfrac{8}{7}=\dfrac{x}{8 \times 7}$
$\Rightarrow x=\dfrac{8 \times 8 \times 7}{7}$
$\therefore x=64$
5. Evaluate $\dfrac{n !}{(nr) !}$, when
(i) $n=6, r=2$
(ii) $n=9, r=5$.
Show Answer
Answer :
(i) When $n=6, r=2, \dfrac{n !}{(nr) !}=\dfrac{6 !}{(62) !}=\dfrac{6 !}{4 !}=\dfrac{6 \times 5 \times 4 !}{4 !}=30$
(ii) When $n=9, r=5, \dfrac{n !}{(nr) !}=\dfrac{9 !}{(95) !}=\dfrac{9 !}{4 !}=\dfrac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 !}$
$=9 \times 8 \times 7 \times 6 \times 5=15120$
6.3.3 Derivation of the formula for ${ }^{n} P_r$
$ { }^{n} P_r=\frac{n !}{(nr) !}, 0 \leq r \leq n $
Let us now go back to the stage where we had determined the following formula:
$$ { }^{n} P_r=n(n1)(n2) \ldots(nr+1) $$
Multiplying numerator and denomirator by $(nr)(nr1) \ldots 3 \times 2 \times 1$, we get
$ { }^{n} P_r=\frac{n(n1)(n2) \ldots(nr+1)(nr)(nr1) \ldots 3 \times 2 \times 1}{(nr)(nr1) \ldots 3 \times 2 \times 1}=\frac{n !}{(nr) !}, $
Thus $\quad \quad \quad$ $ { }^{n} P_r=\frac{n !}{(nr) !} \text{, where } 0<r \leq n $
This is a much more convenient expression for ${ }^{n} P_r$ than the previous one.
In particular, when $r=n,{ }^{n} P_n=\frac{n !}{0 !}=n$ !
Counting permutations is merely counting the number of ways in which some or all objects at a time are rearranged. Arranging no object at all is the same as leaving behind all the objects and we know that there is only one way of doing so. Thus, we can have
$$ { }^{n} P_0=1=\frac{n !}{n !}=\frac{n !}{(n0) !} \quad \quad \quad \quad \quad\quad\quad \ldots(1) $$
Therefore, the formula (1) is applicable for $r=0$ also.
Thus $\quad \quad \quad$ $ { }^{n} P_r=\frac{n !}{(nr) !}, 0 \leq r \leq n $
Theorem 2 The number of permutations of $n$ different objects taken $r$ at a time, where repetition is allowed, is $n^{r}$.
Proof is very similar to that of Theorem 1 and is left for the reader to arrive at.
Here, we are solving some of the problems of the pervious Section using the formula for ${ }^{n} P_r$ to illustrate its usefulness.
In Example 1 , the required number of words $={ }^{4} P_4=4 !=24$. Here repetition is not allowed. If repetition is allowed, the required number of words would be $4^{4}=256$.
The number of 3letter words which can be formed by the letters of the word NUMBER $={ }^{6} P_3=\frac{6 !}{3 !}=4 \times 5 \times 6=120$. Here, in this case also, the repetition is not allowed. If the repetition is allowed, the required number of words would be $6^{3}=216$.
The number of ways in which a Chairman and a ViceChairman can be chosen from amongst a group of 12 persons assuming that one person can not hold more than one position, clearly
$${ }^{12} P_2=\frac{12 !}{10 !}=11 \times 12=132$$.
6.3.4 Permutations when all the objects are not distinct objects
Suppose we have to find the number of ways of rearranging the letters of the word ROOT. In this case, the letters of the word are not all different. There are $2 Os$, which are of the same kind. Let us treat, temporarily, the $2 Os$ as different, say, $O_1$ and $O_2$. The number of permutations of 4different letters, in this case, taken all at a time is 4 !. Consider one of these permutations say, $RO_1 O_2 T$. Corresponding to this permutation, we have 2 ! permutations $RO_1 O_2 T$ and $RO_2 O_1 T$ which will be exactly the same permutation if $O_1$ and $O_2$ are not treated as different, i.e., if $O_1$ and $O_2$ are the same $O$ at both places.Therefore, the required number of permutations $=\frac{4 !}{2 !}=3 \times 4=12$.
Permutations when $O_1, O_2$ are Permutations when $O_1, O_2$ are different. the same $O$.
$\left.\begin{array}{l}\mathrm{RO}_1 \mathrm{O}_2 \mathrm{T} \\ \mathrm{RO}_2 \mathrm{O}_1 \mathrm{T}\end{array}\right] \longrightarrow \quad \mathrm{ROOT}$
$\left.\begin{array}{l}\mathrm{RO}_1 \mathrm{O}_2 \mathrm{T} \\ \mathrm{RO}_2 \mathrm{O}_1 \mathrm{T}\end{array}\right] \longrightarrow \quad \mathrm{ROOT}$
$\left.\begin{array}{l}\mathrm{RO}_1 \mathrm{T} \mathrm{O}_2 \\ \mathrm{RO}_2 \mathrm{T} \mathrm{O}_1\end{array}\right] \longrightarrow \quad \mathrm{ROTO}$
$\left.\begin{array}{l}\mathrm{T} \mathrm{O}_1 \mathrm{RO}_2 \\ \mathrm{TO}_2 \mathrm{R} \mathrm{O}_1\end{array}\right] \longrightarrow \quad \mathrm{TORO}$
$\left.\begin{array}{l}\mathrm{RTO}_1 \mathrm{O}_2 \\ \mathrm{RTO}_2 \mathrm{O}_1\end{array}\right] \longrightarrow \quad \mathrm{RTOO}$
$\left.\begin{array}{l}\mathrm{TRO}_1 \mathrm{O}_2 \\ \mathrm{TRO}_2 \mathrm{O}_1\end{array}\right] \longrightarrow \quad \mathrm{TROO}$
$\left.\begin{array}{l}\mathrm{O}_1 \mathrm{O}_2 \mathrm{RT} \\ \mathrm{O}_2 \mathrm{O}_1 \text { TR }\end{array}\right] \longrightarrow \quad \mathrm{OORT}$
$\left.\begin{array}{c}\mathrm{O}_1 \mathrm{RO}_2 \mathrm{~T} \\ \mathrm{O}_2 \mathrm{RO}_1 \mathrm{~T}\end{array}\right] \longrightarrow \quad \mathrm{OROT}$
$\left.\begin{array}{c}\mathrm{O}_1 \mathrm{TO}_2 \mathrm{R} \\ \mathrm{O}_2 \mathrm{TO}_1 \mathrm{R}\end{array}\right] \longrightarrow \quad \mathrm{OTOR}$
$\left.\begin{array}{lll}\mathrm{O}_1 \mathrm{R} \mathrm{TO}_2 \\ \mathrm{O}_2 \mathrm{R} \mathrm{T} \mathrm{O}_1\end{array}\right] \longrightarrow \quad \mathrm{ORTO}$
$\left.\begin{array}{c}\mathrm{O}_1 \mathrm{TR}_2 \mathrm{O}_2 \\ \mathrm{O}_2 \mathrm{TRO}_1\end{array}\right] \longrightarrow \quad \mathrm{OTRO}$
$\left.\begin{array}{c}\mathrm{O}_1 \mathrm{O}_2 \text { TR } \\ \mathrm{O}_2 \mathrm{O}_1 \text { TR }\end{array}\right] \longrightarrow \quad \mathrm{OOTR}$
Let us now find the number of ways of rearranging the letters of the word INSTITUTE. In this case there are 9 letters, in which I appears 2 times and $T$ appears 3 times.
Temporarily, let us treat these letters different and name them as $I_1, I_2, T_1, T_2, T_3$. The number of permutations of 9 different letters, in this case, taken all at a time is 9 !. Consider one such permutation, say, $I_1 NT_1 SI_2 T_2 UE_3$. Here if $I_1, I_2$ are not same and $T_1, T_2, T_3$ are not same, then $I_1, I_2$ can be arranged in 2 ! ways and $T_1, T_2, T_3$ can be arranged in 3 ! ways. Therefore, $2 ! \times 3$ ! permutations will be just the same permutation corresponding to this chosen permutation $I_1 NT_1 SI_2 T_2 UET_3$. Hence, total number of different permutations will be $\frac{9 !}{2 ! 3 !}$
We can state (without proof) the following theorems:
Theorem 3 The number of permutations of $n$ objects, where $p$ objects are of the same kind and rest are all different $=\frac{n !}{p !}$.
In fact, we have a more general theorem.
Theorem 4 The number of permutations of $n$ objects, where $p_1$ objects are of one kind, $p_2$ are of second kind, …, $p_k$ are of $k^{\text{th }}$ kind and the rest, if any, are of different kind is $\frac{n !}{p_1 ! p_2 ! \ldots p_k !}$.
Example 9 Find the number of permutations of the letters of the word ALLAHABAD.
Solution Here, there are 9 objects (letters) of which there are 4A’s, 2 L’s and rest are all different.Therefore, the required number of arrangements
$$=\frac{9 !}{4 ! 2 !}=\frac{5 \times 6 \times 7 \times 8 \times 9}{2}=7560$$
Example 10 How many 4digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed?
Solution Here order matters for example 1234 and 1324 are two different numbers. Therefore, there will be as many 4 digit numbers as there are permutations of 9 different digits taken 4 at a time.
Therefore, the required 4 digit numbers $={ }^{9} P_4=\frac{9 !}{(94) !}=\frac{9 !}{5 !}=9 \times 8 \times 7 \times 6=3024$.
Example 11 How many numbers lying between 100 and 1000 can be formed with the digits $0,1,2,3,4,5$, if the repetition of the digits is not allowed?
Solution Every number between 100 and 1000 is a 3digit number. We, first, have to count the permutations of 6 digits taken 3 at a time. This number would be ${ }^{6} P_3$. But, these permutations will include those also where 0 is at the 100 ’s place. For example, $092,042, \ldots$, etc are such numbers which are actually 2 digit numbers and hence the number of such numbers has to be subtracted from ${ }^{6} P_3$ to get the required number. To get the number of such numbers, we fix 0 at the 100 ’s place and rearrange the remaining 5 digits taking 2 at a time. This number is ${ }^{5} P_2$. So
The required number $\quad={ }^{6} P_3{ }^{5} P_2=\frac{6 !}{3 !}\frac{5 !}{3 !}$
$ =4 \times 5 \times 64 \times 5=100 $
Example 12 Find the value of $n$ such that (i) ${ }^{n} P_5=42{ }^{n} P_3, n>4$ (ii) $\frac{{ }^{n} P_4}{{ }^{n1} P_4}=\frac{5}{3}, n>4$
Solution (i) Given that
$$ { }^{n} \mathrm{P} _{5}=42{ }^{n} \mathrm{P} _{3} $$
or $\quad \quad \quad \quad$ $n(n1)(n2)(n3)(n4)=42 n(n1)(n2)$
Since $\quad \quad \quad \quad$ $n>4 \quad$ so $n(n1)(n2) \neq 0$
Therefore, by dividing both sides by $n(n1)(n2)$, we get
$$\begin{array}{ll} {} & (n3(n4)=42 \\ \text{or}\quad\quad & n^{2}7 n30=0 \\ \text{or} & n^{2}10 n+3 n30 \\ \text{or} & (n10)(n+3)=0 \\ \text{or} & n10=0 \text{ or } \quad n+3=0 \\ \text{or} & n=10 \quad \text{ or } \quad n=3 \end{array}$$
As $n$ cannot be negative, so $n=10$.
(ii) Given that $\frac{{ }^{n} P _4}{{ }^{n1} P _4}=\frac{5}{3}$
Therefore $\quad \quad 3 n(n1)(n2)(n3)=5(n1)(n2)(n3)(n4)$
or $\quad \quad3 n=5(n4) \quad[$ as $(n1)(n2)(n3) \neq 0, n>4]$
or $\quad \quad n=10$.
Example 13 Find $r$, if $5{ }^{4} P_r=6{ }^{5} P _{r1}$.
Solution We have $5{ }^{4} P_r=6{ }^{5} P _{r1}$
or $ \quad\quad5 \times \frac{4 !}{(4r) !}=6 \times \frac{5 !}{(5r+1) !} $
or $ \quad\quad\frac{5 !}{(4r) !}=\frac{6 \times 5 !}{(5r+1)(5r)(5r1) !} $
or $\quad(6r)(5r)=6$
or $\quad r^{2}11 r+24=0$
or $\quad r^{2}8 r3 r+24=0$
or $\quad(r8)(r3)=0$
or $\quad r=8$ or $r=3$.
Hence $\quad r=8,3$.
Example 14 Find the number of different 8letter arrangements that can be made from the letters of the word DAUGHTER so that
(i) all vowels occur together
(ii) all vowels do not occur together.
Solution (i) There are 8 different letters in the word DAUGHTER, in which there are 3 vowels, namely, A, U and E. Since the vowels have to occur together, we can for the time being, assume them as a single object (AUE). This single object together with 5 remaining letters (objects) will be counted as 6 objects. Then we count permutations of these 6 objects taken all at a time. This number would be ${ }^{6} P_6=6$ !. Corresponding to each of these permutations, we shall have 3 ! permutations of the three vowels A, U, E taken all at a time . Hence, by the multiplication principle the required number of permutations $=6 ! \times 3 !=4320$.
(ii) If we have to count those permutations in which all vowels are never together, we first have to find all possible arrangments of 8 letters taken all at a time, which can be done in 8 ! ways. Then, we have to subtract from this number, the number of permutations in which the vowels are always together.
Therefore, the required number
$$ \begin{aligned} 8 !6 ! \times 3 ! & =6 !(7 \times 86) \\ & =2 \times 6 !(283) \\ & =50 \times 6 !=50 \times 720=36000 \end{aligned} $$
Example 15 In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable?
Solution Total number of discs are $4+3+2=9$. Out of 9 discs, 4 are of the first kind (red), 3 are of the second kind (yellow) and 2 are of the third kind (green).
Therefore, the number of arrangements $\frac{9 !}{4 ! 3 ! 2 !}=1260$.
Example 16 Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements,
(i) do the words start with $P$
(ii) do all the vowels always occur together
(iii) do the vowels never occur together
(iv) do the words begin with I and end in P?
Solution There are 12 letters, of which $N$ appears 3 times, $E$ appears 4 times and D appears 2 times and the rest are all different. Therefore
The required number of arrangements $=\frac{12 !}{3 ! 4 ! 2 !}=1663200$
(i) Let us fix $P$ at the extreme left position, we, then, count the arrangements of the remaining 11 letters. Therefore, the required number of words starting with $P$
$$ =\frac{11 !}{3 ! 2 ! 4 !}=138600 $$
(ii) There are 5 vowels in the given word, which are 4 Es and 1 I. Since, they have to always occur together, we treat them as a single object EEEEI for the time being. This single object together with 7 remaining objects will account for 8 objects. These 8 objects, in which there are $3 Ns$ and $2 Ds$, can be rearranged in $8 !$ $\overline{3 ! 2 !}$ ways. Corresponding to each of these arrangements, the 5 vowels E, E, E, E and I can be rearranged in $\frac{5 !}{4 !}$ ways. Therefore, by multiplication principle, the required number of arrangements
$$ =\frac{8 !}{3 ! 2 !} \times \frac{5 !}{4 !}=16800 $$
(iii) The required number of arrangements $=$ the total number of arrangements (without any restriction)  the number of arrangements where all the vowels occur together.
$$ =166320016800=1646400 $$
(iv) Let us fix I and P at the extreme ends (I at the left end and P at the right end). We are left with 10 letters.
Hence, the required number of arrangements $ =\frac{10 !}{3 ! 2 ! 4 !}=12600 $
EXERCISE 6.3
1. How many 3digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
Show Answer
Answer :
3digit numbers have to be formed using the digits 1 to 9 .
Here, the order of the digits matters.
Therefore, there will be as many 3digit numbers as there are permutations of 9 different digits taken 3 at a time.
Therefore, required number of 3digit numbers
$ ={ }^{9} P_3=\dfrac{9 !}{(93) !}=\dfrac{9 !}{6 !} $
$=\dfrac{9 \times 8 \times 7 \times 6 !}{6 !}=9 \times 8 \times 7=504$
2. How many 4digit numbers are there with no digit repeated?
Show Answer
Answer :
There are four places for the digits in a 4digit number: thousands, hundreds, tens, and units. We can fill these places one by one following these constraints:
1. Thousands place: The digit cannot be zero, because then the number would become a 3digit number. So, there are 9 choices $(1, 2, 3, 4, 5, 6, 7, 8, 9)$ for the thousands place.
2. Hundreds place: Since no digit can be repeated, there are only $9$ digits remaining to fill the hundreds place.
3. Tens place: After filling the thousands and hundreds places, there are 8 digits remaining to fill the tens place.
4. Units place: Finally, there are $7$ digits remaining to fill the units place.
By the fundamental counting principle, the total number of unique 4digit numbers possible is the product of the number of choices at each place:
Total numbers = Choices (thousands) × Choices (hundreds) × Choices (tens) × Choices (units) $= 9 × 9 × 8 × 7$
Therefore, there are $4536$ ($9$ x $504)$ different 4digit numbers that can be formed with no digits repeated.
3. How many 3digit even numbers can be made using the digits $1,2,3,4,6,7$, if no digit is repeated?
Show Answer
Answer :
3digit even numbers are to be formed using the given six digits, 1, 2, 3, 4, 6, and 7, without repeating the digits.
Then, units digits can be filled in 3 ways by any of the digits, 2, 4, or 6 .
Since the digits cannot be repeated in the 3digit numbers and units place is already occupied with a digit (which is even), the hundreds and tens place is to be filled by the remaining 5 digits.
Therefore, the number of ways in which hundreds and tens place can be filled with the remaining 5 digits is the permutation of 5 different digits taken 2 at a time.
Number of ways of filling hundreds and tens place
$ ={ }^{5} P_2=\dfrac{5 !}{(52) !}=\dfrac{5 !}{3 !} $
$=\dfrac{5 \times 4 \times 3 !}{3 !}=20$
Thus, by multiplication principle, the required number of 3digit numbers is
$3 \times 20=60$
4. Find the number of 4digit numbers that can be formed using the digits 1,2,3,4, 5 if no digit is repeated. How many of these will be even?
Show Answer
Answer :
4digit numbers are to be formed using the digits, 1, 2, 3, 4, and 5 .
There will be as many 4digit numbers as there are permutations of 5 different digits taken 4 at a time.
Therefore, required number of 4 digit numbers $=$
$ { }^{5} P_4=\dfrac{5 !}{(54) !}=\dfrac{5 !}{1 !} $
$=1 \times 2 \times 3 \times 4 \times 5=120$
Among the 4digit numbers formed by using the digits, 1, 2, 3, 4, 5, even numbers end with either 2 or 4 .
The number of ways in which units place is filled with digits is 2 .
Since the digits are not repeated and the units place is already occupied with a digit (which is even), the remaining places are to be filled by the remaining 4 digits.
Therefore, the number of ways in which the remaining places can be filled is the permutation of 4 different digits taken 3 at a time.
Number of ways of filling the remaining places
$ ={ }^{4} P_3=\dfrac{4 !}{(43) !}=\dfrac{4 !}{1 !} $
$=4 \times 3 \times 2 \times 1=24$
Thus, by multiplication principle, the required number of even numbers is
$24 \times 2=48$
5. From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?
Show Answer
Answer :
From a committee of 8 persons, a chairman and a vice chairman are to be chosen in such a way that one person cannot hold more than one position.
Here, the number of ways of choosing a chairman and a vice chairman is the permutation of 8 different objects taken 2 at a time.
Thus, required number of ways $=$
$ ={ }^{8} P_2=\dfrac{8 !}{(82) !}=\dfrac{8 !}{6 !}=\dfrac{8 \times 7 \times 6 !}{6 !}=8 \times 7=56 $
6. Find $n$ if ${ }^{n1} P_3:{ }^{n} P_4=1: 9$.
Show Answer
Answer :
$ \begin{aligned} & { }^{n1} P_3:{ }^{n} P_4=1: 9 \\ & \Rightarrow \dfrac{{ }^{n1} P_3}{{ }^{n} P_4}=\dfrac{1}{9} \\ & \Rightarrow \dfrac{[\dfrac{(n1) !}{(n13) !}]}{[\dfrac{n !}{(n4) !}]}=\dfrac{1}{9} \\ & \Rightarrow \dfrac{(n1) !}{(n4) !} \times \dfrac{(n4) !}{n !}=\dfrac{1}{9} \\ & \Rightarrow \dfrac{(n1) !}{n \times(n1) !}=\dfrac{1}{9} \\ & \Rightarrow \dfrac{1}{n}=\dfrac{1}{9} \\ & \therefore n=9 \end{aligned} $
7. Find $r$ if (i) ${ }^{5} P_r=2{ }^{6} P _{r1}$ (ii) ${ }^{5} P_r={ }^{6} P _{r1}$.
Show Answer
Answer:
(i)
$ \begin{aligned} & { }^{5} P_r=2^{6} P _{r1} \\ & \Rightarrow \dfrac{5 !}{(5r) !}=2 \times \dfrac{6 !}{(6r+1) !} \\ & \Rightarrow \dfrac{5 !}{(5r) !}=\dfrac{2 \times 6 !}{(7r) !} \\ & \Rightarrow \dfrac{5 !}{(5r) !}=\dfrac{2 \times 6 \times 5 !}{(7r)(6r)(5r) !} \\ & \Rightarrow 1=\dfrac{2 \times 6}{(7r)(6r)} \\ & \Rightarrow(7r)(6r)=12 \end{aligned} $
$ \begin{aligned} & \Rightarrow 426 r7 r+r^{2}=12 \\ & \Rightarrow r^{2}13 r+30=0 \\ & \Rightarrow r^{2}3 r10 r+30=0 \\ & \Rightarrow r(r3)10(r3)=0 \\ & \Rightarrow(r3)(r10)=0 \\ & \Rightarrow(r3)=0 \text{ or }(r10)=0 \\ & \Rightarrow r=3 \text{ or } r=10 \end{aligned} $
It is known that,
$ { }^{n} P_r=\dfrac{n !}{(nr) !} \text{, where } 0 \leq r \leq n $
$\therefore 0 \leq r \leq 5$
Hence, $r \neq 10$
$\therefore r=3$
(ii)
$ \begin{aligned} & { }^{5} P_r={ }^{6} P _{r1} \\ & \Rightarrow \dfrac{5 !}{(5r) !}=\dfrac{6 !}{(6r+1) !} \\ & \Rightarrow \dfrac{5 !}{(5r) !}=\dfrac{6 \times 5 !}{(7r) !} \\ & \Rightarrow \dfrac{1}{(5r) !}=\dfrac{6}{(7r)(6r)(5r) !} \\ & \Rightarrow 1=\dfrac{6}{(7r)(6r)} \\ & \Rightarrow(7r)(6r)=6 \\ & \Rightarrow 427 r6 r+r^{2}6=0 \\ & \Rightarrow r^{2}13 r+36=0 \\ & \Rightarrow r^{2}4 r9 r+36=0 \\ & \Rightarrow r(r4)9(r4)=0 \\ & \Rightarrow(r4)(r9)=0 \\ & \Rightarrow(r4)=0 \text{ or }(r9)=0 \\ & \Rightarrow r=4 \text{ or } r=9 \end{aligned} $
It is known that,
$ { }^{n} P_r=\dfrac{n !}{(nr) !} \text{, where } 0 \leq r \leq n $
Hence, $r \neq 9$
$\therefore r=4$
8. How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
Show Answer
Answer :
There are 8 different letters in the word EQUATION.
Therefore, the number of words that can be formed using all the letters of the word EQUATION, using each letter exactly once, is the number of permutations of 8 different objects taken 8 at a time, which is ${ }^{8} P_8=8$ !.
Thus, required number of words that can be formed $=8 !=40320$
9. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.
(i) 4 letters are used at a time,
(ii) all letters are used at a time,
(iii) all letters are used but first letter is a vowel?
Show Answer
Answer :
There are 6 different letters in the word MONDAY.
(i) Number of 4letter words that can be formed from the letters of the word MONDAY, without repetition of letters, is the number of permutations of 6 different objects taken 4 at a time, which is ${ }^{6} P_4$.
Thus, required number of words that can be formed using 4 letters at a time is
$ { }^{6} P_4=\dfrac{6 !}{(64) !}=\dfrac{6 !}{2 !}=\dfrac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}=6 \times 5 \times 4 \times 3=360 $
(ii) Number of words that can be formed by using all the letters of the word MONDAY at a time is the number of permutations of 6 different objects taken 6 at a time, which is ${ }^{6} P_6=6$ !
Thus, required number of words that can be formed when all letters are used at a time $=6 !=6 \times 5 \times 4 \times 3 \times 2 \times 1=$ 720
(iii) In the given word, there are 2 different vowels, which have to occupy the rightmost place of the words formed. This can be done only in 2 ways.
Since the letters cannot be repeated and the rightmost place is already occupied with a letter (which is a vowel), the remaining five places are to be filled by the remaining 5 letters. This can be done in 5 ! ways.
Thus, in this case, required number of words that can be formed is
$5 ! \times 2=120 \times 2=240$
10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
Show Answer
Answer :
In the given word MISSISSIPPI, I appears 4 times, S appears 4 times, P appears 2 times, and M appears just once. Therefore, number of distinct permutations of the letters in the given word
$ \begin{aligned} & =\dfrac{11 !}{4 ! 4 ! 2 !} \\ & =\dfrac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 ! \times 4 \times 3 \times 2 \times 1 \times 2 \times 1} \\ & =\dfrac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1 \times 2 \times 1} \\ & =34650 \end{aligned} $
There are 4 Is in the given word. When they occur together, they are treated as a single object IIII for the time being. This single object together with the remaining 7 objects will account for 8 objects.
These 8 objects in which there are 4 Ss and 2 Ps can be arranged in $\dfrac{8 !}{4 ! 2 !}$ ways i.e., 840 ways.
Number of arrangements where all Is occur together $=840$
Thus, number of distinct permutations of the letters in MISSISSIPPI in which four Is do not come together $=34650$$840=33810$
11. In how many ways can the letters of the word PERMUTATIONS be arranged if the
(i) words start with $P$ and end with $S$,
(ii) vowels are all together,
(iii) there are always 4 letters between $P$ and $S$ ?
Show Answer
Answer :
In the word PERMUTATIONS, there are $2 Ts$ and all the other letters appear only once.
(i) If $P$ and $S$ are fixed at the extreme ends ( $P$ at the left end and $S$ at the right end), then 10 letters are left.
Hence, in this case, required number of arrangements
$ =\dfrac{10 !}{2 !}=1814400 $
(ii) There are 5 vowels in the given word, each appearing only once.
Since they have to always occur together, they are treated as a single object for the time being. This single object together with the remaining 7 objects will account for 8 objects. These 8 objects in which there are 2 Ts can be
arranged in $\dfrac{8 !}{2 !}$ ways .
Corresponding to each of these arrangements, the 5 different vowels can be arranged in 5 ! ways.
Therefore, by multiplication principle, required number of arrangements in this case $=\dfrac{8 !}{2 !} \times 5 !=2419200$
(iii) The letters have to be arranged in such a way that there are always 4 letters between $P$ and $S$.
Therefore, in a way, the places of $P$ and $S$ are fixed. The remaining 10 letters in which there are $2 Ts$ can be arranged in $\dfrac{10 !}{2 !}$ ways
Also, the letters $P$ and $S$ can be placed such that there are 4 letters between them in $2 \times 7=14$ ways.
Therefore, by multiplication principle, required number of arrangements in this case
$ =\dfrac{10 !}{2 !} \times 14=25401600 $
6.4 Combinations
Let us now assume that there is a group of 3 lawn tennis players $X, Y, Z$. A team consisting of 2 players is to be formed. In how many ways can we do so? Is the team of $X$ and $Y$ different from the team of $Y$ and $X$ ? Here, order is not important. In fact, there are only 3 possible ways in which the team could be constructed.
These are XY, YZ and ZX (Fig 6.3).
Here, each selection is called a combination of 3 different objects taken 2 at a time. In a combination, the order is not important.
Now consider some more illustrations.
Twelve persons meet in a room and each shakes hand with all the others. How do we determine the number of hand shakes. $X$ shaking hands with $Y$ and $Y$ with $X$ will not be two different hand shakes. Here, order is not important. There will be as many hand shakes as there are combinations of 12 different things taken 2 at a time.
Seven points lie on a circle. How many chords can be drawn by joining these points pairwise? There will be as many chords as there are combinations of 7 different things taken 2 at a time.
Now, we obtain the formula for finding the number of combinations of $n$ different objects taken $r$ at a time, denoted by ${ }^{n} C_r$.
Suppose we have 4 different objects A, B, C and D. Taking 2 at a time, if we have to make combinations, these will be $AB, AC, AD, BC, BD, CD$. Here, $AB$ and $BA$ are the same combination as order does not alter the combination. This is why we have not included BA, CA, DA, CB, DB and DC in this list. There are as many as 6 combinations of 4 different objects taken 2 at a time, i.e., ${ }^{4} C_2=6$.
Corresponding to each combination in the list, we can arrive at 2! permutations as 2 objects in each combination can be rearranged in 2! ways. Hence, the number of permutations $={ }^{4} C_2 \times 2$!.
On the other hand, the number of permutations of 4 different things taken 2 at a time $={ }^{4} P_2$.
Therefore $\quad { }^{4} P_2= ^{4} C_2 \times 2 !$ or $\frac{4 !}{(42) ! 2 !}= ^{4} C_2$
Now, let us suppose that we have 5 different objects A, B, C, D, E. Taking 3 at a time, if we have to make combinations, these will be $ABC, ABD, ABE, BCD, BCE$, $CDE, ACE, ACD, ADE, BDE$. Corresponding to each of these ${ }^{5} C_3$ combinations, there are 3! permutations, because, the three objects in each combination can be rearranged in 3! ways. Therefore, the total of permutations $={ }^{5} C_3 \times 3$!
Therefore $\quad{ }^{5} P_3={ }^{5} C_3 \times 3$ ! or $\quad \frac{5 !}{(53) ! 3 !}={ }^{5} C_3$
These examples suggest the following theorem showing relationship between permutaion and combination:
Theorem 5 ${ }^{n} P_r={ }^{n} C_r r!, 0<r \leq n$.
Proof Corresponding to each combination of ${ }^{n} C_r$, we have $r$! permutations, because $r$ objects in every combination can be rearranged in $r$ ! ways.
Hence, the total number of permutations of $n$ different things taken $r$ at a time is ${ }^{n} C_r \times r !$. On the other hand, it is ${ }^{n} P_r$.
Thus $ { }^{n} P_r={ }^{n} C_r \times r !, 0<r \leq n $
Remarks 1. From above $\frac{n !}{(nr) !}={ }^{n} C_r \times r !$, i.e., $\quad{ }^{n} C_r=\frac{n !}{r !(nr) !}$.
In particular, if $r=n,{ }^{n} C_n=\frac{n !}{n ! 0 !}=1$.
2. We define ${ }^{n} C_0=1$, i.e., the number of combinations of $n$ different things taken nothing at all is considered to be 1 . Counting combinations is merely counting the number of ways in which some or all objects at a time are selected. Selecting nothing at all is the same as leaving behind all the objects and we know that there is only one way of doing so. This way we define ${ }^{n} C_0=1$.
3. As $\frac{n !}{0 !(n0) !}=1={ }^{n} C_0$, the formula ${ }^{n} C_r=\frac{n !}{r !(nr) !}$ is applicable for $r=0$ also.
Hence $ { }^{n} C_r=\frac{n !}{r !(nr) !}, 0 \leq r \leq n . $
4. $\quad{ }^{n} C _{nr}=\frac{n !}{(nr) !(n(nr)) !}=\frac{n !}{(nr) ! r !}={ }^{n} C_r$, i.e., selecting $r$ objects out of $n$ objects is same as rejecting $(nr)$ objects.
5. $\quad{ }^{n} C_a={ }^{n} C_b \Rightarrow a=b$ or $a=nb$, i.e., $n=a+b$
Theorem 6 ${ }^{n} C_r+{ }^{n} C _{r1}={ }^{n+1} C_r$
Proof We have $\quad{ }^{n} C_r+{ }^{n} C _{r1}=\frac{n !}{r !(nr) !}+\frac{n !}{(r1) !(nr+1) !}$
$$ \begin{aligned} & =\frac{n !}{r \times(r1) !(nr) !}+\frac{n !}{(r1) !(nr+1)(nr) !} \\ & =\frac{n !}{(r1) !(nr) !}[\frac{1}{r}+\frac{1}{nr+1}] \\ & =\frac{n !}{(r1) !(nr) !} \times \frac{nr+1+r}{r(nr+1)}=\frac{(n+1) !}{r !(n+1r) !}={ }^{n+1} C_r \end{aligned} $$
Example 17 If ${ }^{n} C_9={ }^{n} C_8$, find ${ }^{n} C _{17}$.
Solution We have ${ }^{n} C_9={ }^{n} C_8$
i.e., $\quad \frac{n !}{9 !(n9) !}=\frac{n !}{(n8) ! 8 !}$
or $ \quad\quad\quad \frac{1}{9}=\frac{1}{n8} \text{ or } n8=9 \text{ or } n=17 $
Therefore $ \quad \quad { }^{n} C _{17}={ }^{17} C _{17}=1 . $
Example 18 A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?
Solution Here, order does not matter. Therefore, we need to count combinations. There will be as many committees as there are combinations of 5 different persons taken 3 at a time. Hence, the required number of ways $={ }^{5} C_3=\frac{5 !}{3 ! 2 !}=\frac{4 \times 5}{2}=10$.
Now, 1 man can be selected from 2 men in ${ }^{2} C_1$ ways and 2 women can be selected from 3 women in ${ }^{3} C_2$ ways. Therefore, the required number of committees
$$ ={ }^{2} C_1 \times{ }^{3} C_2=\frac{2 !}{1 ! 1 !} \times \frac{3 !}{2 ! 1 !}=6 $$
Example 19 What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these
(i) four cards are of the same suit,
(ii) four cards belong to four different suits,
(iii) are face cards,
(iv) two are red cards and two are black cards,
(v) cards are of the same colour?
Solution There will be as many ways of choosing 4 cards from 52 cards as there are combinations of 52 different things, taken 4 at a time. Therefore The required number of ways
$$ ={ }^{52} \mathrm{C}_{4}=\frac{52 !}{4 ! 48 !}=\frac{49 \times 50 \times 51 \times 52}{2 \times 3 \times 4}=270725 $$
(i) There are four suits: diamond, club, spade, heart and there are 13 cards of each suit. Therefore, there are ${ }^{13} C_4$ ways of choosing 4 diamonds. Similarly, there are ${ }^{13} C_4$ ways of choosing 4 clubs, ${ }^{13} C_4$ ways of choosing 4 spades and ${ }^{13} C_4$ ways of choosing 4 hearts. Therefore
The required number of ways $={ }^{13} C_4+{ }^{13} C_4+{ }^{13} C_4+{ }^{13} C_4$.
$$ =4 \times \frac{13 !}{4 ! 9 !}=2860 $$
(ii) There are13 cards in each suit. Therefore, there are ${ }^{13} C_1$ ways of choosing 1 card from 13 cards of diamond, ${ }^{13} C_1$ ways of choosing 1 card from 13 cards of hearts, ${ }^{13} C_1$ ways of choosing 1 card from 13 cards of clubs, ${ }^{13} C_1$ ways of choosing 1 card from 13 cards of spades. Hence, by multiplication principle, the required number of ways
$$ ={ }^{13} C_1 \times{ }^{13} C_1 \times{ }^{13} C_1 \times{ }^{13} C_1=13^{4} $$
(iii) There are 12 face cards and 4 are to be selected out of these 12 cards. This can be done in ${ }^{12} C_4$ ways. Therefore, the required number of ways $=\frac{12 !}{4 ! 8 !}=495$.
(iv) There are 26 red cards and 26 black cards. Therefore, the required number of ways
$$ \begin{aligned} & ={ }^{26} \mathrm{C} _{2} \times{ }^{26} \mathrm{C} _{2} \\ & =\frac{26 !}{2 ! 24 !}^{2}=(325)^{2}=105625 \end{aligned} $$
(v) 4 red cards can be selected out of 26 red cards in ${ }^{26} C_4$ ways. 4 black cards can be selected out of 26 black cards in ${ }^{26} C_4$ ways.
Therefore, the required number of ways $={ }^{26} C_4+{ }^{26} C_4$ $$ =2 \times \frac{26 !}{4 ! 22 !}=29900 . $$
EXERCISE 6.4
1. If ${ }^{n} C_8={ }^{n} C_2$, find ${ }^{n} C_2$.
Show Answer
Answer :
It is known that, ${ }^{n} C_a={ }^{n} C_b \Rightarrow a=b$ or $n=a+b$
Therefore,
$ \begin{aligned} & { }^{n} C_8={ }^{n} C_2 \Rightarrow n=8+2=10 \\ & \therefore{ }^{n} C_2={ }^{10} C_2=\dfrac{10 !}{2 !(102) !}=\dfrac{10 !}{2 ! 8 !}=\dfrac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !}=45 \end{aligned} $
2. Determine $n$ if
(i) ${ }^{2 n} C_3:{ }^{n} C_3=12: 1$
(ii) ${ }^{2 n} C_3:{ }^{n} C_3=11: 1$
Show Answer
Answer :
(i)
$ \begin{aligned} & \dfrac{{ }^{2 n} C_3}{{ }^{n} C_3}=\dfrac{12}{1} \\ & \Rightarrow \dfrac{(2 n) !}{3 !(2 n3) !} \times \dfrac{3 !(n3) !}{n !}=\dfrac{12}{1} \\ & \Rightarrow \dfrac{(2 n)(2 n1)(2 n2)(2 n3) !}{(2 n3) !} \times \dfrac{(n3) !}{n(n1)(n2)(n3) !}=12 \\ & \Rightarrow \dfrac{2(2 n1)(2 n2)}{(n1)(n2)}=12 \\ & \Rightarrow \dfrac{4(2 n1)(n1)}{(n1)(n2)}=12 \\ & \Rightarrow \dfrac{(2 n1)}{(n2)}=3 \\ & \Rightarrow 2 n1=3(n2) \\ & \Rightarrow 2 n1=3 n6 \\ & \Rightarrow 3 n2 n=1+6 \\ & \Rightarrow n=5 \end{aligned} $
(ii)
$\dfrac{{ }^{2 n} C_3}{{ }^{n} C_3}=\dfrac{11}{1}$
$\Rightarrow \dfrac{(2 n) !}{3 !(2 n3) !} \times \dfrac{3 !(n3) !}{n !}=11$
$\Rightarrow \dfrac{(2 n)(2 n1)(2 n2)(2 n3) !}{(2 n3) !} \times \dfrac{(n3) !}{n(n1)(n2)(n3) !}=11$
$\Rightarrow \dfrac{2(2 n1)(2 n2)}{(n1)(n2)}=11$
$\Rightarrow \dfrac{4(2 n1)(n1)}{(n1)(n2)}=11$
$\Rightarrow \dfrac{4(2 n1)}{n2}=11$
$\Rightarrow 4(2 n1)=11(n2)$
$\Rightarrow 8 n4=1 \ln 22$
$\Rightarrow 1 \ln 8 n=4+22$
$\Rightarrow 3 n=18$
$\Rightarrow n=6$
3. How many chords can be drawn through 21 points on a circle?
Show Answer
Answer :
For drawing one chord on a circle, only 2 points are required.
To know the number of chords that can be drawn through the given 21 points on a circle, the number of combinations have to be counted.
Therefore, there will be as many chords as there are combinations of 21 points taken 2 at a time.
Thus, required number of chords $=$
$ { }^{21} C_2=\dfrac{21 !}{2 !(212) !}=\dfrac{21 !}{2 ! 19 !}=\dfrac{21 \times 20}{2}=210 $
4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Show Answer
Answer :
A team of 3 boys and 3 girls is to be selected from 5 boys and 4 girls.
3 boys can be selected from 5 boys in ${ }^{5} C_3$ ways.
3 girls can be selected from 4 girls in ${ }^{4} C_3$ ways.
Therefore, by multiplication principle, number of ways in which a team of 3 boys and 3 girls can be
selected
$ ={ }^{5} C_3 \times{ }^{4} C_3=\dfrac{5 !}{3 ! 2 !} \times \dfrac{4 !}{3 ! 1 !} $
$=\dfrac{5 \times 4 \times 3 !}{3 ! \times 2} \times \dfrac{4 \times 3 !}{3 !}$
$=10 \times 4=40$
5. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Show Answer
Answer :
There are a total of 6 red balls, 5 white balls, and 5 blue balls.
9 balls have to be selected in such a way that each selection consists of 3 balls of each colour.
Here,
3 balls can be selected from 6 red balls in ${ }^{6} C_3$ ways.
3 balls can be selected from 5 white balls in ${ }^{5} C_3$ ways.
3 balls can be selected from 5 blue balls in ${ }^{5} C_3$ ways.
Thus, by multiplication principle, required number of ways of selecting 9 balls
$ \begin{aligned} & ={ }^{6} C_3 \times{ }^{5} C_3 \times{ }^{5} C_3=\dfrac{6 !}{3 ! 3 !} \times \dfrac{5 !}{3 ! 2 !} \times \dfrac{5 !}{3 ! 2 !} \\ & =\dfrac{6 \times 5 \times 4 \times 3 !}{3 ! \times 3 \times 2} \times \dfrac{5 \times 4 \times 3 !}{3 ! \times 2 \times 1} \times \dfrac{5 \times 4 \times 3 !}{3 ! \times 2 \times 1} \\ & =20 \times 10 \times 10=2000 \end{aligned} $
6. Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Show Answer
Answer :
In a deck of 52 cards, there are 4 aces. A combination of 5 cards have to be made in which there is exactly one ace.
Then, one ace can be selected in ${ }^{4} C_1$ ways and the remaining 4 cards can be selected out of the 48 cards in ${ }^{48} C_4$ ways.
$ \begin{aligned} & ={ }^{48} C_4 \times{ }^{4} C_1=\dfrac{48 !}{4 ! 44 !} \times \dfrac{4 !}{1 ! 3 !} \\ & =\dfrac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1} \times 4 \end{aligned} $
Thus, by multiplication principle, required number of 5 card combinations $=778320$
7. In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?
Show Answer
Answer :
Out of 17 players, 5 players are bowlers.
A cricket team of 11 players is to be selected in such a way that there are exactly 4 bowlers.
4 bowlers can be selected in ${ }^{5} C_4$ ways and the remaining 7 players can be selected out of the 12 players in ${ }^{12} C_7$ ways.
Thus, by multiplication principle, required number of ways of selecting cricket
$ \text{ team }={ }^{5} C_4 \times{ }^{12} C_7=\dfrac{5 !}{4 ! 1 !} \times \dfrac{12 !}{7 ! 5 !}=5 \times \dfrac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1}=3960 $
8. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
Show Answer
Answer :
There are 5 black and 6 red balls in the bag.
2 black balls can be selected out of 5 black balls in ${ }^{5} C_2$ ways and 3 red balls can be selected out of 6 red balls in ${ }^{6} C_3$ ways.
Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red
balls $={ }^{5} C_2 \times{ }^{6} C_3=\dfrac{5 !}{2 ! 3 !} \times \dfrac{6 !}{3 ! 3 !}=\dfrac{5 \times 4}{2} \times \dfrac{6 \times 5 \times 4}{3 \times 2 \times 1}=10 \times 20=200$
9. In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
Show Answer
Answer :
There are 9 courses available out of which, 2 specific courses are compulsory for every student.
Therefore, every student has to choose 3 courses out of the remaining 7 courses. This can be chosen in ${ }^{7} C_3$ ways. Thus, required number of ways of choosing the programme
$ ={ }^{7} C_3=\dfrac{7 !}{3 ! 4 !}=\dfrac{7 \times 6 \times 5 \times 4 !}{3 \times 2 \times 1 \times 4 !}=35 $
Miscellaneous Examples
Example 20 How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE ?
Solution In the word INVOLUTE, there are 4 vowels, namely, I,O,E,Uand 4 consonants, namely, $N, V, L$ and $T$.
The number of ways of selecting 3 vowels out of $4={ }^{4} C_3=4$.
The number of ways of selecting 2 consonants out of $4={ }^{4} C_2=6$.
Therefore, the number of combinations of 3 vowels and 2 consonants is $4 \times 6=24$.
Now, each of these 24 combinations has 5 letters which can be arranged among themselves in 5 ! ways. Therefore, the required number of different words is $24 \times 5 !=2880$.
Example 21 A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (i) no girl ? (ii) at least one boy and one girl ? (iii) at least 3 girls ?
Solution (i) Since, the team will not include any girl, therefore, only boys are to be selected. 5 boys out of 7 boys can be selected in ${ }^{7} C_5$ ways. Therefore, the required number of ways $={ }^{7} C_5=\frac{7 !}{5 ! 2 !}=\frac{6 \times 7}{2}=21$
(ii) Since, at least one boy and one girl are to be there in every team. Therefore, the team can consist of
(a) 1 boy and 4 girls
(b) 2 boys and 3 girls
(c) 3 boys and 2 girls
(d) 4 boys and 1 girl.
1 boy and 4 girls can be selected in ${ }^{7} C_1 \times{ }^{4} C_4$ ways.
2 boys and 3 girls can be selected in ${ }^{7} C_2 \times{ }^{4} C_3$ ways.
3 boys and 2 girls can be selected in ${ }^{7} C_3 \times{ }^{4} C_2$ ways.
4 boys and 1 girl can be selected in ${ }^{7} C_4 \times{ }^{4} C_1$ ways.
Therefore, the required number of ways
$ \begin{aligned} & ={ }^{7} C_1 \times{ }^{4} C_4+{ }^{7} C_2 \times{ }^{4} C_3+{ }^{7} C_3 \times{ }^{4} C_2+{ }^{7} C_4 \times{ }^{4} C_1 \\ & =7+84+210+140=441 \end{aligned} $
(iii) Since, the team has to consist of at least 3 girls, the team can consist of (a) 3 girls and 2 boys, or (b) 4 girls and 1 boy.
Note that the team cannot have all 5 girls, because, the group has only 4 girls.
3 girls and 2 boys can be selected in ${ }^{4} C_3 \times{ }^{7} C_2$ ways.
4 girls and 1 boy can be selected in ${ }^{4} C_4 \times{ }^{7} C_1$ ways.
Therefore, the required number of ways
$ ={ }^{4} C_3 \times{ }^{7} C_2+{ }^{4} C_4 \times{ }^{7} C_1=84+7=91 $
Example 22 Find the number of words with or without meaning which can be made using all the letters of the word AGAIN. If these words are written as in a dictionary, what will be the $50^{\text{th }}$ word?
Solution There are 5 letters in the word AGAIN, in which A appears 2 times. Therefore, the required number of words $=\frac{5 !}{2 !}=60$.
To get the number of words starting with A, we fix the letter A at the extreme left position, we then rearrange the remaining 4 letters taken all at a time. There will be as many arrangements of these 4 letters taken 4 at a time as there are permutations of 4 different things taken 4 at a time. Hence, the number of words starting with $A=4 !=24$. Then, starting with $G$, the number of words $=\frac{4 !}{2 !}=12$ as after placing $G$ at the extreme left position, we are left with the letters A, A, I and N. Similarly, there are 12 words starting with the next letter I. Total number of words so far obtained $=24+12+12=48$. The $49^{\text{th }}$ word is NAAGI. The $50^{\text{th }}$ word is NAAIG.
Example 23 How many numbers greater than 1000000 can be formed by using the digits $1,2,0,2,4,2,4$ ?
Solution Since, 1000000 is a 7digit number and the number of digits to be used is also 7. Therefore, the numbers to be counted will be 7digit only. Also, the numbers have to be greater than 1000000 , so they can begin either with 1,2 or 4 .
The number of numbers beginning with $1=\frac{6 !}{3 ! 2 !}=\frac{4 \times 5 \times 6}{2}=60$, as when 1 is fixed at the extreme left position, the remaining digits to be rearranged will be $0,2,2,2$, 4,4 , in which there are $3,2 s$ and $2,4 s$.
Total numbers begining with 2 $=\frac{6 !}{2 ! 2 !}=\frac{3 \times 4 \times 5 \times 6}{2}=180$
and total numbers begining with $4=\frac{6 !}{3 !}=4 \times 5 \times 6=120$
Therefore, the required number of numbers $=60+180+120=360$.
Alternative Method
The number of 7digit arrangements, clearly, $\frac{7 !}{3 ! 2 !}=420$. But, this will include those numbers also, which have 0 at the extreme left position. The number of such arrangements $\frac{6 !}{3 ! 2 !}$ (by fixing 0 at the extreme left position) $=60$. Therefore, the required number of numbers $=42060=360$.
Note  If one or more than one digits given in the list is repeated, it will be understood that in any number, the digits can be used as many times as is given in the list, e.g., in the above example 1 and 0 can be used only once whereas 2 and 4 can be used 3 times and 2 times, respectively.
Example 24 In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together?
Solution Let us first seat the 5 girls. This can be done in 5 ! ways. For each such arrangement, the three boys can be seated only at the cross marked places.
$$ \times G \times G \times G \times G \times G \times . $$
There are 6 cross marked places and the three boys can be seated in ${ }^{6} P_3$ ways. Hence, by multiplication principle, the total number of ways
$$ \begin{aligned} & =5 ! \times{ }^{6} P_3=5 ! \times \frac{6 !}{3 !} \\ & =4 \times 5 \times 2 \times 3 \times 4 \times 5 \times 6=14400 . \end{aligned} $$
Miscellaneous Exercise on Chapter 6
1. How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
Show Answer
Answer :
In the word DAUGHTER, there are 3 vowels namely, A, U, and E, and 5 consonants namely, D, G, H, T, and R.
Number of ways of selecting 2 vowels out of 3 vowels $={ }^{3} C_2=3$
Number of ways of selecting 3 consonants out of 5 consonants $={ }^{5} C_3=10$
Therefore, number of combinations of 2 vowels and 3 consonants $=3 \times 10=30$
Each of these 30 combinations of 2 vowels and 3 consonants can be arranged among themselves in 5 ! ways.
Hence, required number of different words $=30 \times 5 !=3600$
2. How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
Show Answer
Answer :
In the word EQUATION, there are 5 vowels, namely, $A, E, I, O$, and $U$, and 3 consonants, namely, Q, $T$, and $N$. Since all the vowels and consonants have to occur together, both (AEIOU) and (QTN) can be assumed as single objects. Then, the permutations of these 2 objects taken all at a time are counted. This number would be ${ }^{2} P_2=2$ ! Corresponding to each of these permutations, there are 5 ! permutations of the five vowels taken all at a time and 3 ! permutations of the 3 consonants taken all at a time.
Hence, by multiplication principle, required number of words $=2 ! \times 5 ! \times 3 !$
$=1440$
3. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:
(i) exactly 3 girls ?
(ii) atleast 3 girls ?
(iii) atmost 3 girls ?
Show Answer
Answer :
A committee of 7 has to be formed from 9 boys and 4 girls.
i. Since exactly 3 girls are to be there in every committee, each committee must consist of $(73)$ boys only.
Thus, in this case, required number of ways $=$
$ { }^{4} C_3 \times{ }^{9} C_4=\dfrac{4 !}{3 ! 1 !} \times \dfrac{9 !}{4 ! 5 !} $
$=4 \times \dfrac{9 \times 8 \times 7 \times 6 \times 5 !}{4 \times 3 \times 2 \times 1 \times 5 !}$
$=504$
(ii) Since at least 3 girls are to be there in every committee, the committee can consist of
(a) 3 girls and 4 boys or (b) 4 girls and 3 boys
3 girls and 4 boys can be selected in ${ }^{4} C_3 \times{ }^{9} C_4$ ways.
4 girls and 3 boys can be selected in ${ }^{4} C_4 \times{ }^{9} C_3$ ways.
Therefore, in this case, required number of ways $={ }^{4} C_3 \times{ }^{9} C_4+{ }^{4} C_4 \times{ }^{9} C_3$ $=504+84=588$
(iii) Since atmost 3 girls are to be there in every committee, the committee can consist of
(a) 3 girls and 4 boys (b) 2 girls and 5 boys
(c) 1 girl and 6 boys (d) No girl and 7 boys
3 girls and 4 boys can be selected in ${ }^{4} C_3 \times{ }^{9} C_4$ ways.
2 girls and 5 boys can be selected in ${ }^{4} C_2 \times{ }^{9} C_5$ ways.
1 girl and 6 boys can be selected in ${ }^{4} C_1 \times{ }^{9} C_6$ ways.
No girl and 7 boys can be selected in ${ }^{4} C_0 \times{ }^{9} C_7$ ways.
Therefore, in this case, required number of ways
$={ }^{4} C_3 \times{ }^{9} C_4+{ }^{4} C_2 \times{ }^{9} C_5+{ }^{4} C_1 \times{ }^{9} C_6+{ }^{4} C_0 \times{ }^{9} C_7$
$=\dfrac{4 !}{3 ! 1 !} \times \dfrac{9 !}{4 ! 5 !}+\dfrac{4 !}{2 ! 2 !} \times \dfrac{9 !}{5 ! 4 !}+\dfrac{4 !}{1 ! 3 !} \times \dfrac{9 !}{6 ! 3 !}+\dfrac{4 !}{0 ! 4 !} \times \dfrac{9 !}{7 ! 2 !}$
$=504+756+336+36$
$=1632$
4. If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with $E$ ?
Show Answer
Answer :
In the given word EXAMINATION, there are 11 letters out of which, $A, I$, and $N$ appear 2 times and all the other letters appear only once.
The words that will be listed before the words starting with $E$ in a dictionary will be the words that start with $A$ only.
Therefore, to get the number of words starting with A, the letter A is fixed at the extreme left position, and then the remaining 10 letters taken all at a time are rearranged.
Since there are 2 Is and $2 Ns$ in the remaining 10 letters,
Number of words starting with $A=\dfrac{10 !}{2 ! 2 !}=907200$
Thus, the required numbers of words is 907200 .
5. How many 6digit numbers can be formed from the digits $0,1,3,5,7$ and 9 which are divisible by 10 and no digit is repeated?
Show Answer
Answer :
A number is divisible by 10 if its units digits is 0 .
Therefore, 0 is fixed at the units place.
Therefore, there will be as many ways as there are ways of filling 5 vacant places
$ \begin{array}{cccccc} \hline & & & & & 0\\ \hline \end{array} $
by the remaining 5 digits (i.e., 1, 3, 5, 7 and 9 ).
The 5 vacant places can be filled in 5 ! ways.
Hence, required number of 6digit numbers $=5 !=120$
6. The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?
Show Answer
Answer :
2 different vowels and 2 different consonants are to be selected from the English alphabet.
Since there are 5 vowels in the English alphabet, number of ways of selecting 2 different vowels from the alphabet
$ ={ }^{5} C_2=\dfrac{5 !}{2 ! 3 !}=10 $
Since there are 21 consonants in the English alphabet, number of ways of selecting 2 different consonants from the alphabet
$ ={ }^{21} C_2=\dfrac{21 !}{2 ! 19 !}=210 $
Therefore, number of combinations of 2 different vowels and 2 different consonants $=10 \times 210=2100$
Each of these 2100 combinations has 4 letters, which can be arranged among themselves in 4! ways.
Therefore, required number of words $=2100 \times 4 !=50400$
7. In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?
Show Answer
Answer :
It is given that the question paper consists of 12 questions divided into two parts Part I and Part II, containing 5 and 7 questions, respectively.
A student has to attempt 8 questions, selecting at least 3 from each part.
This can be done as follows.
(a) 3 questions from part I and 5 questions from part II
(b) 4 questions from part I and 4 questions from part II
(c) 5 questions from part I and 3 questions from part II
3 questions from part I and 5 questions from part II can be selected in ${ }^{5} C_3 \times{ }^{7} C_5$ ways.
4 questions from part I and 4 questions from part II can be selected in ${ }^{5} C_4 \times{ }^{7} C_4$ ways.
5 questions from part I and 3 questions from part II can be selected in ${ }^{5} C_5 \times{ }^{7} C_3$ ways.
Thus, required number of ways of selecting questions
$ \begin{aligned} & ={ }^{5} C_3 \times{ }^{7} C_5+{ }^{5} C_4 \times{ }^{7} C_4+{ }^{5} C_5 \times{ }^{7} C_3 \\ & =\dfrac{5 !}{2 ! 3 !} \times \dfrac{7 !}{2 ! 5 !}+\dfrac{5 !}{4 ! 1 !} \times \dfrac{7 !}{4 ! 3 !}+\dfrac{5 !}{5 ! 0 !} \times \dfrac{7 !}{3 ! 4 !} \\ & =210+175+35=420 \end{aligned} $
8. Determine the number of 5card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
Show Answer
Answer :
From a deck of 52 cards, 5 card combinations have to be made in such a way that in each selection of 5 cards, there is exactly one king.
In a deck of 52 cards, there are 4 kings.
1 king can be selected out of 4 kings in ${ }^{4} C_1$ ways.
4 cards out of the remaining 48 cards can be selected in ${ }^{48} C_4$ ways.
Thus, the required number of 5card combinations is ${ }^{4} C_1 \times{ }^{48} C_4$.
9. It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?
Show Answer
Answer :
5 men and 4 women are to be seated in a row such that the women occupy the even places.
The 5 men can be seated in 5 ! ways. For each arrangement, the 4 women can be seated only at the cross marked places (so that women occupy the even places).
$ M \times M \times M \times M \times M $
Therefore, the women can be seated in 4! ways.
Thus, possible number of arrangements $=4 ! \times 5 !=24 \times 120=2880$
10. From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?
Show Answer
Answer :
From the class of 25 students, 10 are to be chosen for an excursion party.
Since there are 3 students who decide that either all of them will join or none of them will join, there are two cases.
Case I: All the three students join.
Then, the remaining 7 students can be chosen from the remaining 22 students in ${ }^{22} C_7$ ways.
Case II: None of the three students join.
Then, 10 students can be chosen from the remaining 22 students in ${ }^{22} C _{10}$ ways.
Thus, required number of ways of choosing the excursion party is ${ }^{22} C_7+{ }^{22} C _{10}$.
11. In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?
Show Answer
Answer :
In the given word ASSASSINATION, the letter A appears 3 times, $S$ appears 4 times, I appears 2 times, $N$ appears 2 times, and all the other letters appear only once.
Since all the words have to be arranged in such a way that all the Ss are together, SSSS is treated as a single object for the time being. This single object together with the remaining 9 objects will account for 10 objects.
These 10 objects in which there are 3 As, 2 ls, and 2 Ns can be arranged in $\dfrac{10 !}{3 ! 2 ! 2 !}$ ways.
Thus, required number of ways of arranging the letters of the given word
$=\dfrac{10 !}{3 ! 2 ! 2 !}=151200$
Summary

Fundamental principle of counting If an event can occur in $m$ different ways, following which another event can occur in $n$ different ways, then the total number of occurrence of the events in the given order is $m \times n$.

The number of permutations of $n$ different things taken $r$ at a time, where repetition is not allowed, is denoted by ${ }^{n} P_r$ and is given by ${ }^{n} P_r=\frac{n !}{(nr) !}$, where $0 \leq r \leq n$.

$n !=1 \times 2 \times 3 \times \ldots \times n$

$n !=n \times(n1)$ !

The number of permutations of $n$ different things, taken $r$ at a time, where repeatition is allowed, is $n^{r}$.

$\diamond$ The number of permutations of $n$ objects taken all at a time, where $p_1$ objects are of first kind, $p_2$ objects are of the second kind, $\ldots, p_k$ objects are of the $k^{\text{th }}$ kind and rest, if any, are all different is $\frac{n !}{p_1 ! p_2 ! \ldots p_k !}$.

The number of combinations of $n$ different things taken $r$ at a time, denoted by ${ }^{n} C_r$, is given by ${ }^{n} C_r==\frac{n !}{r !(nr) !}, 0 \leq r \leq n$.
Historical Note
The concepts of permutations and combinations can be traced back to the advent of Jainism in India and perhaps even earlier. The credit, however, goes to the Jains who treated its subject matter as a selfcontained topic in mathematics, under the name Vikalpa.
Among the Jains, Mahavira, (around 850) is perhaps the world’s first mathematician credited with providing the general formulae for permutations and combinations.
In the 6th century B.C., Sushruta, in his medicinal work, Sushruta Samhita, asserts that 63 combinations can be made out of 6 different tastes, taken one at a time, two at a time, etc. Pingala, a Sanskrit scholar around third century B.C., gives the method of determining the number of combinations of a given number of letters, taken one at a time, two at a time, etc. in his work Chhanda Sutra. Bhaskaracharya (born 1114) treated the subject matter of permutations and combinations under the name Anka Pasha in his famous work Lilavati. In addition to the general formulae for ${ }^{n} C_r$ and ${ }^{n} P_r$ already provided by Mahavira, Bhaskaracharya gives several important theorems and results concerning the subject.
Outside India, the subject matter of permutations and combinations had its humble beginnings in China in the famous book IKing (Book of changes). It is difficult to give the approximate time of this work, since in 213 B.C., the emperor had ordered all books and manuscripts in the country to be burnt which fortunately was not completely carried out. Greeks and later Latin writers also did some scattered work on the theory of permutations and combinations.
Some Arabic and Hebrew writers used the concepts of permutations and combinations in studying astronomy. Rabbi ben Ezra, for instance, determined the number of combinations of known planets taken two at a time, three at a time and so on. This was around 1140. It appears that Rabbi ben Ezra did not know the formula for ${ }^{n} C_r$. However, he was aware that ${ }^{n} C_r={ }^{n} C _{nr}$ for specific values $n$ and $r$. In 1321, Levi Ben Gerson, another Hebrew writer came up with the formulae for ${ }^{n} P_r,{ }^{n} P_n$ and the general formula for ${ }^{n} C_r$.
The first book which gives a complete treatment of the subject matter of permutations and combinations is Ars Conjectandi written by a Swiss, Jacob Bernoulli (1654  1705), posthumously published in 1713. This book contains essentially the theory of permutations and combinations as is known today.