PERMUTATION AND COMBINATIONS - 2 (Simple Applications on ${}^n P_r$ and ${}^n C_r$)
Important Results
1. Sum of digits in the unit place of all numbers formed by $a _{1}, a _{2} \ldots \ldots \ldots a _{n}$ taken all at a time is given by $(n-1) !\left(a _{1}+a _{2}+\ldots . . a _{n}\right)$ if repetition of digits is not allowe(d).
2. Sum of all the numbers which can be formed using the digits $\mathrm{a} _{1}, \mathrm{a} _{2} \ldots \ldots \ldots \mathrm{a} _{\mathrm{n}}$ (repetition not allowed)
$ \begin{aligned} & =\quad(\mathrm{n}-1) !\left(\mathrm{a} _{1}+\mathrm{a} _{2}+\ldots \ldots+\mathrm{a} _{\mathrm{n}}\right) \frac{\left(10^{\mathrm{n}}-1\right)}{9} \\ & =\quad(\mathrm{n}-1) !(\text { sum of digits) }(\underbrace{1 \ldots \ldots 1} _{\mathrm{n} \text { times }}) \end{aligned} $
Number of integral solutions of linear equations and unequations (Multinomial Theorem)
3. (i). Total number of non negative integral solutions of $\mathrm{x} _{1}+\mathrm{x} _{2}+\ldots \ldots \mathrm{x} _{\mathrm{r}}=\mathrm{n}$ is ${ }^{\mathrm{n}+\mathrm{r}-1} \mathrm{C} _{\mathrm{r}-1}$ Total number of positive integral solutions of $\mathrm{x} _{1}+\mathrm{x} _{2}+\ldots \ldots . \mathrm{x} _{\mathrm{r}}=\mathrm{n}$ is ${ }^{\mathrm{n}-1} \mathrm{C} _{\mathrm{r}-1}$
(ii). In order to solve inequations of the form $\mathrm{x} _{1}+\mathrm{x} _{2}+\ldots . .+\mathrm{x} _{\mathrm{r}} \leq \mathrm{n}$, we introduce artificial (dummy) variable $\mathrm{x} _{\mathrm{r}+1}$ such that $\mathrm{x} _{1}+\mathrm{x} _{2}+\ldots \ldots .+\mathrm{x} _{\mathrm{r}}+\mathrm{x} _{\mathrm{r}+1}=\mathrm{n}$ where $\mathrm{x} _{\mathrm{r}+1} \geq 0$.
Number of solutions of this equation are same as the number of solutions of inequation $\mathrm{x} _{1}+\mathrm{x} _{2}+\ldots . \mathrm{x} _{\mathrm{r}} \leq \mathrm{n}$.
(iii). Number of solutions of $\alpha+2 \beta+3 \gamma+\ldots . .+\mathrm{q} \theta=\mathrm{n}$ is
$\left\{\begin{array}{l}\text { coefficient of } x^{n} \text { in }(1-x)^{-1}\left(1-x^{2}\right)^{-1}\left(1-x^{3}\right)^{-1} \ldots \ldots .\left(1-x^{q}\right)^{-1} \text { if zero is included } \\ \text { coefficient of } x^{n} \text { in } x^{1+2+\ldots \ldots+q}(1-x)^{-1}\left(1-x^{2}\right)^{-1} \ldots \ldots .\left(1-x^{q}\right)^{-1} \text { if zero is not included }\end{array}\right.$
Number and sum of divisors
4. Let $\mathrm{N}=\mathrm{a}^{\mathrm{p}} \mathrm{b}^{\mathrm{q}} \mathrm{c}^{\mathrm{r}}$ where $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are primes & $\mathrm{p}, \mathrm{q}, \mathrm{r} \in \mathrm{Z}^{+}$.
(i). Number of divisors of $\mathrm{N}=(\mathrm{p}+1)(\mathrm{q}+1)(\mathrm{r}+1)$
Sum of divisors of $\mathrm{N}=\left(1+\mathrm{a}+\mathrm{a}^{2}+\ldots . \mathrm{a}^{\mathrm{p}}\right)\left(1+\mathrm{b}+\mathrm{b}^{2}+\ldots \mathrm{b}^{\mathrm{q}}\right)\left(1+\mathrm{c}+\mathrm{c}^{2}+\ldots . \mathrm{c}^{\mathrm{r}}\right)$
(ii). Number of ways in which $\mathrm{N}$ can be resolved as a product of two factors is
$ \left\{\begin{array}{l} \frac{1}{2}(a+1)(b+1)(c+1) \text { if } N \text { is not perfect square } \\ \left\{\frac{1}{2}(a+1)(b+1)(c+1)+1\right\} \text { if } N \text { is a perfect square } \end{array}\right\} $
5. Number of ways in which a composite number $\mathrm{N}$ can be resolved into two factors which are relatively prime (or co prime) to each other is $2^{\mathrm{n}-1}$ where $\mathrm{n}$ is the number of different prime factors is $\mathrm{N}$.
Divisibility
6. Condition for divisibility of a number
A number abcde will be divisible
1. by 4 if $2 \mathrm{~d}+\mathrm{e}$ is divisible by 4
2. by 8 if $4 \mathrm{c}+2 \mathrm{~d}+\mathrm{e}$ is divisible by 8
3. by 3 if $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}$ is divisible by 3
4. by 9 if $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}$ is divisible by 9
5. by 5 if $\mathrm{e}=0$ or 5
6. by 11 if $\underbrace{\mathrm{a}+\mathrm{c}+\mathrm{e}} _{\begin{array}{c}\text { Sum of digits } \\ \text { at odd places }\end{array}}-\underset{\begin{array}{c}\text { sur of digit } \\ \text { at even places }\end{array}}{\mathrm{b}+\mathrm{d}}$ in divisible by 11
7. by 6 if $\mathrm{e}=$ even and $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}$ is divisible by 3
8. by 18 if $\mathrm{e}=$ even and $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}$ is divisible by 9
Solved examples
1. Ten different letters are printed round a circle. The number of different ways in which we can select three letters so that no two of them are consecutive is
(a). 26
(b). 50
(c). 56
(d). 72
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Solution:
Number of selections is
${ }^{10} \mathrm{C} _{3}-10-10 \cdot{ }^{6} \mathrm{C} _{2}=50$ ways as total ways of selection is ${ }^{10} \mathrm{C} _{3}$ (number of selections is ${ }^{\mathrm{n}} \mathrm{C} _{3}-\mathrm{n}-\mathrm{n}^{\mathrm{n} .4} \mathrm{C} _{2}$ )
Number of ways when three are consecutive is 10
Number of ways when two are consecutive is $10 .{ }^{6} \mathrm{C} _{2}$. Subtract these two cases from the total number of ways.
Answer: b
2. The numberof triangles whose vertices are the vertices of an octagon but none of whose sides happen to come from the sides of octagon is
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Solution:
(a). 24
(b). 52
(c). 48
(d). 16
Proceding in a similar way as in solved example 1 , we have number of selection as ${ }^{8} \mathrm{C} _{2}-8-8 .{ }^{4} \mathrm{C} _{2}=16$ (number of selections is ${ }^{n} \mathrm{C} _{3}-\mathrm{n}-\mathrm{n} . .^{\mathrm{n} 4} \mathrm{C} _{2}$ )
Answer: d
3. The maximum number of points in which 4 circles and 4 straight lines intersect is
(a). 26
(b). 50
(c). 56
(d). 72
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Solution:
Maximum number of points of line - line intersection $={ }^{4} \mathrm{C} _{2}=6$
Maximum number of points circle - circle intersction is ${ }^{4} \mathrm{P} _{2}=12$
Maximum number of points of line circle intersection is $(2 \times 4) \times 4=32$
$\therefore$ total number of points of intersection is $6+12+32=50$.
Answer: b
4. In a plane two families of lines are given by $y=x+r$ and $(y=-x+p)$ where $r \in\{0,1,2,3,4\}$ and $p \in\{0,1,2, \ldots \ldots, 9\}$. The number of squares of diagonals of length 3 units formed by these lines is
(a). 36
(b). 24
(c). 20
(d). none of these
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Solution:
It means we have to select two lines from each family in such a way that there is a gap of 2 lines between the selected lines. First pair can be selected in two ways and second pair can be selected in seven ways. Hence, number of squares selected is $7 \times 2=14$
Answer: d
5. The total number of words that can be made using letters of the word CALCULATE so that each word starts and ends with a consonant is
(a). $\frac{5.7 !}{2}$
(b). $\frac{3.7 !}{2}$
(c). $2.7 !$
(d). none of these
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Solution:
Consonants | Vowels |
---|---|
CLT | AUE |
CL | A |
Arrangements can be as follows.
C | $\frac{7 !}{2 ! 2 !}$ | C |
C | $\frac{7 !}{2 !}$ | L |
L | $\frac{7 !}{2 !}$ | C |
C | $\frac{7 !}{2 ! 2 !}$ | T |
T | $\frac{7 !}{2 ! 2 !}$ | C |
C | $\frac{7 !}{2 ! 2 !}$ | T |
L | $\frac{7 !}{2 ! 2 !}$ | T |
T | $\frac{7 !}{2 ! 2 !}$ | T |
$\therefore \quad$ Total number of arrangements can be $\frac{7 !}{2 !}\left(\frac{1}{2}+1+1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)$
$ =\frac{5.7 !}{2} $
6. The number of 5 digit numbers of different digints in which middle digits is the largest is
(a). $ \sum _{\mathrm{n}=4}^{9} \mathrm{P} _{4}$
(b). $33(3 !)$
(c). $30(3 !)$
(d). none of these
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Solution: Fix the middle digit. Number of arrangements is
$ \left({ }^{4} \mathrm{P} _{4}-{ }^{3} \mathrm{P} _{3}\right)+\left({ }^{5} \mathrm{P} _{4}-{ }^{4} \mathrm{P} _{3}\right)+\ldots \ldots \ldots+\left({ }^{9} \mathrm{P} _{4}-{ }^{8} \mathrm{P} _{3}\right) $
Answer: d
Practice questions
1. Last digit of $(1 !+2 !+\ldots . .+2005 !)^{500}$ is
(a). 9
(b). 2
(c). 7
(d). 1
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Answer: (d)2. Number of integral solutions of $x+y+z=0$ with $x \geq-5, y \geq-5, z \geq-5$ is
(a). 134
(b). 136
(c). 138
(d). 140
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Answer: (b)3. Let $\mathrm{x} _{1}, \mathrm{x} _{2} \ldots, \mathrm{x} _{\mathrm{k}}$ be divisors of positive integer $\mathrm{n}$ (excluding 1 & $\mathrm{n}$ ). If $\mathrm{x} _{1}+\mathrm{x} _{2}+\ldots .+\mathrm{x} _{\mathrm{k}}=75$, then $\sum _{\mathrm{i}=1}^{\mathrm{k}} \frac{1}{\mathrm{x} _{\mathrm{i}}}$ is equal to
(a). $\frac{75}{\mathrm{n}^{2}}$
(b). $\frac{75}{n}$
(c). $\frac{75}{\mathrm{k}}$
(d). none of these
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Answer: (b)4. The total number of ways in which $\mathrm{n}^{2}$ number of identical balls can be put in $\mathrm{n}$ numbered boxes $(1,2,3, \ldots, n)$ such that $\mathrm{i}^{\text {th }}$ box contains at least $\mathrm{i}$ number of balls is
(a). ${ }^{n^{2}} C _{n-1}$
(b). ${ }^{n^{2}-1} C _{n-1}$
(c). $ \frac{n^{2}+n-2}{2} C _{n-1}$
(d). none of these
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Answer: (c)5. Total number of positive integral solutions of $15<\mathrm{x} _{1}+\mathrm{x} _{2}+\mathrm{x} _{3} \leq 20$ is
(a). 685
(b). 785
(c). 1125
(d). none of these
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Answer: (a)6. If $n$ is selected from the set $\{1,2,3 \ldots .10\}$ and the number $2^{n}+3^{n}+5^{n}$ is forme(d). Total number of ways of selecting $n$ so that the formed number is divisible by 4 is equal to
(a). 50
(b). 49
(c). 48
(d). none of these
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Answer: (b)7. $A$ is a set containing $n$ different elements. A subset $P$ of $A$ is chosen. The set $A$ is reconstructed by replacing the elements of $\mathrm{P}$. A subset $\mathrm{Q}$ of $\mathrm{A}$ is again chosen. The number of ways of choosing $\mathrm{P}$ and $\mathrm{Q}$ so that $\mathrm{P} \cap \mathrm{Q}$ contains exactly two elements is
(a). ${ }^{n} C _{3} \times 2^{n}$
(b). ${ }^{n} C _{2} \times 3^{n-2}$
(c). $3^{\mathrm{n}-2}$
(d). none of these
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Answer: (b)8. The number of three digit numbers of the form $x y z$ such that $x<y$ and $z \leq y$ is
(a). 276
(b). 285
(c). 240
(d). 244
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Answer: (a)9. Number of ordered triplets $(x, y, z)$ such that $x, y, z$ are primes and $x^{y}+1=z$ is
(a). 0
(b). 1
(c). 3
(d). none of these
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Answer: (b)10. Read the passage and answer the following questions:-
Suppose a lot of $n$ objects having $n _{1}$ objects one kind, $n _{2}$ objects of second kind, $n _{3}$ objects of third kind,…, $\mathrm{n} _{\mathrm{k}}$ objects of $\mathrm{k}^{\text {th }}$ kind satisfying the condition $\mathrm{n} _{1}+\mathrm{n} _{2}+\ldots+\mathrm{n} _{\mathrm{k}}=\mathrm{n}$, then the number of possible arrangements / permutations of $m$ objects out of this lot is the coefficient of $\mathrm{x}^{\mathrm{m}}$ in the expansion of $m ! \prod\left\{\sum _{\lambda=0}^{n i} \frac{x \lambda}{\lambda i}\right\}$
i. The number of permutations of the letters of the word AGAIN taken three at a time is
(a). 48
(b). 24
(c). 36
(d). 33
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Answer: (d)ii. The number of permutations of the letters of the word EXAMINATION taken 4 at a time is
(a). 136
(b). 2454
(c). 2266
(d). none of these
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Answer: (b)iii. The number of permutations of the letters of the word EXERCISES taken 5 at a time is
(a). 2250
(b). 30240
(c). 226960
(d). none of these
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Answer: (a)iv. The number of ways in which an arrangement of 4 letters of the word PROPORTION can be made is
(a). 700
(b). 750
(c). 758
(d). none of these
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Answer: (c)v. The number of permutations of the letters of the word SURITI taken 4 at a time is
(a). 360
(b). 240
(c). 216
(d). none of these