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

Show Answer

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

Show Answer

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

Show Answer

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

Show Answer

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

Show Answer

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

Show Answer

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

Show Answer 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

Show Answer 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

Show Answer 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

Show Answer 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

Show Answer 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

Show Answer 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

Show Answer 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

Show Answer 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

Show Answer 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

Show Answer 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

Show Answer 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

Show Answer 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

Show Answer 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

Show Answer Answer: (d)


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