Important Results

1. Sum of digits in the unit place of all numbers formed by $\mathrm{a} _{1}, \mathrm{a} _{2} \ldots \ldots \ldots \mathrm{a} _{\mathrm{n}}$ taken all at a time is given by $(n-1) !\left(a _{1}+a _{2}+\ldots \ldots a _{n}\right)$ if repetition of digits is not allowed.

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)

$\begin{aligned} & =(n-1)!\left(a_1+a_2+\ldots \ldots+a_n\right) \frac{\left(10^n-1\right)}{9} \\ & =(n-1)!(\text { sum of digits })(\underset{n \text { times }}{11 \ldots \ldots 1}) \end{aligned}$

3. Number of whole number solutions $\left(x _{i} \geq 0 \forall i\right)$ (non-negative) of $x _{1}+x _{2}+\ldots . . x _{r}=n$ is ${ }^{n-1} C _{r-1}$ Number of solutions of $\alpha+2 \beta+3 \gamma+\ldots .+\mathrm{q} \theta=\mathrm{n}$ is $\left\{\begin{array}{l}\text { coefficient of } x^{n} \text { in }(1-x)^{-1}\left(1-x^{2}\right)^{-1}\left(1-x^{3}\right)^{-1} \ldots \ldots\left(1-x^{q}\right)^{-1} \text { if zero is included } \\ \text { coefficient of } x^{n} \text { in } x^{1+2+\ldots .+q}(1-x)^{-1}\left(1-x^{2}\right)^{-1} \ldots \ldots .\left(1-x^{q}\right)^{-1} \text { if zero is not included }\end{array}\right.$

4. Number and sum of divisors

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. $\quad$ Number of divisors of $\mathrm{N}=(\mathrm{p}+1)(\mathrm{q}+1)(\mathrm{r}+1)$ Sum of divisors of $\mathrm{N}=\left(1+\mathrm{a}+\mathrm{a}^{2}+\ldots \mathrm{a}^{\mathrm{p}}\right)\left(1+\mathrm{b}+\mathrm{b}^{2}+\ldots\right.$ b) $\left(1+\mathrm{c}+\mathrm{c}^{2}+\ldots . \mathrm{c}^{\mathrm{r}}\right)$

ii. Number of ways in which $\mathrm{N}$ can be resolved as a product of two factors is

$\left\{\begin{array}{l} \dfrac{1}{2}(a+1)(b+1)(c+1) \text { if } N \text { is not perfect square } \\ \left\{\dfrac{1}{2}(a+1)(b+1)(c+1)+1\right\} \text { if } N \text { is a perfect square } \end{array}\right\}$

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 $\mathrm{N}$.

Multinomial Theorem:

Coefficient of $x^{r}$ is $(1-x)^{-n}={ }^{n+r-1} C _{r}$. Number of ways of making a selection from $m+n+p=N$ things where $p$ are alike of one kind, $m$ alike of second kind and $n$ alike of third kind taken $r$ at a time is given by coefficient of $x^{r}$ is expansion of $\left(1+x+x^2+\ldots \ldots . x^m\right)\left(1+x+x^2+\ldots \ldots . x^n\right)\left(1+x+x^2+\ldots \ldots . x^p\right)$

Example: Number of selection of 4 letter words from the letters for the ward PROPROTION is

$\{\mathrm{PP}, \mathrm{RR}, \mathrm{OOO}, \mathrm{T}, \mathrm{I}, \mathrm{N}\}$

Coefficient of $\mathrm{x}^{4}$ is $\left(1+\mathrm{x}+\mathrm{x}^{2}\right)\left(1+\mathrm{x}+\mathrm{x}^{2}\right)\left(1+\mathrm{x}+\mathrm{x}^{2}+\mathrm{x}^{3}\right)(1+\mathrm{x})(1+\mathrm{x})(1+\mathrm{x})$

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 $\underbrace{\mathrm{a}+\mathrm{c}+\mathrm{e}} _{\substack{\text { Sum of tigits } \ \text { at odd places }}}-\underbrace{\mathrm{b}+\mathrm{d}} _{\substack{\text { sut of didit } \ \text { at ven places }}}$ is 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. The number of divisors of $(6 !)^{3 !}$ is

(a) 364

(b) 9100

(c) 2275

(d) 75

2. The number of ways in which three district number in AP can be selected from $1,2,3——–, 24$ is

(a) 132

(b) 572

(c) 264

(d) 150

3. If $x, y, z$ are integers and $x \geq 0, y \geq 1, z \geq 2$ and $x+y+z=15$, then the number of ordered triplets $(\mathrm{x}, \mathrm{y}, \mathrm{z})$ is

(a) 91

(b) 455

(c) ${ }^{17} \mathrm{C} _{2}$

(d) none of these

4. $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ are odd natural numbers such that $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=20$, then number of quadrapulets $(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d})$ is

(a) 165

(b) 455

(c) 310

(d) 255

5. The number of positive integral solutions of $x+y+z \leq 10$ is ____________________.

Exercise:

1. Number of divisors of the form $(4 n+2) ;(n \geq 0)$ of the integer 240 is

(a) 4

(b) 8

(c) 10

(d) 13

2. If $\mathrm{r}, \mathrm{s}, \mathrm{t}$ are prime numbers and $\mathrm{p}, \mathrm{q}$ are the positive integers such that $\mathrm{LCM}$ of $\mathrm{p}, \mathrm{q}$ is $\mathrm{r}^{2} \mathrm{~s}^{4} \mathrm{t}^{2}$, then the number of ordered pairs $(p, q)$ is

(a) 252

(b) 254

(c) 225

(d) 224

3. The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1,2 and 3 only, is

(a) 55

(b) 66

(c) 77

(d) 88

4. Let $\mathrm{n}$ and $\mathrm{k}$ be positive integers such that $\mathrm{n} \geq{ }^{\mathrm{k}+1} \mathrm{C} _{2}$. The number of solutions $\left(\mathrm{x} _{1}, \mathrm{x} _{2}, \ldots . \mathrm{x} _{\mathrm{k}}\right)$; $\mathrm{x} _{1} \geq 1 \geq \mathrm{x} _{2} \geq 2, \ldots \ldots . \mathrm{x} _{\mathrm{k}} \geq \mathrm{k}$ all integers satisfying $\mathrm{x} _{1}+\mathrm{x} _{2}+\ldots \ldots+\mathrm{x} _{\mathrm{k}}=\mathrm{n}$ is

(a) $\quad{ }^{n-\dfrac{k}{2}} C _{k}$

(b) $\quad{ }^{n-1-\dfrac{k}{2}} C _{k}$

(c) $\quad{ }^{n-1-\dfrac{k}{2}} C _{k-1}$

(d) $\quad{ }^{n+1-\dfrac{k}{2}} C _{k-1}$

5. The number of divisors of the form $4 \mathrm{n}+1, \mathrm{n} \geq 0$ of the number $10^{10} 11^{11} 13^{13}$ is

(a) 750

(b) 840

(c) 924

(d) 1024

6. The number of positive integer solution of the equation $\left[\dfrac{x}{99}\right]=\left[\dfrac{x}{101}\right]$ is

(a) 2500

(b) 2499

(c) 1729

(d) 1440

7. Let $\mathrm{N}$ be natural number. If its first digit (form the left) is deleted, it gets reduced to $\dfrac{\mathrm{IV}}{57}$. The sum of all the digits of $\mathrm{N}$ is

(a) 15

(b) 18

(c) 24

(d) 30

8. The number of positive integral pairs $(\mathrm{x}, \mathrm{y})$ such that $\dfrac{1}{\mathrm{x}}+\dfrac{1}{\mathrm{y}}=\dfrac{1}{2007}, \mathrm{x}<\mathrm{y}$ is

(a) 5

(b) 6

(c) 7

(d) 8

9. The number of ordered triplets of positive integers which satisfy the inequality $15 \leq x+y+z \leq 45$ is

(a) ${ }^{45} \mathrm{C} _{2}-{ }^{14} \mathrm{C} _{2}$

(b) ${ }^{45} \mathrm{C} _{3}-{ }^{14} \mathrm{C} _{3}$

(c) ${ }^{46} \mathrm{C} _{3}{ }^{-15} \mathrm{C} _{3}$

(d) none of these

10.* Match the following:

11. Read the passage and answer the following questions

Five balls are to be placed in 3 boxes. Each can hold all the five balls. In how many ways can we place the balls so that no box remains empty, when

i. Balls and boxes are all different

(a) 150

(b) 6

(c) 50

(d) 2

ii. balls are identical but boxes are different

(a) 150

(b) 6

(c) 50

(d) 2

iii. balls are different but boxes are identical

(a) 150

(b) 6

(c) 50

(d) 2

iv. balls as well as boxes are identical.

(a) 150

(b) 6

(c) 50

(d) 2