CAT 2025 Lesson : Permutations & Combinations - Identical Elements

A bag contains $8$ identical red balls. In how many different ways can
(I) John pick balls from the bag?
(II) Josie pick balls from the bag, such that she picks at least $1$ ball?

Solution

Case I: John can pick $0, 1, 2, 3, 4, 5, 6, 7$ or $8$ balls.
Number of ways $= 8 + 1 = 9$

Case II: Josie can pick $1, 2, 3, 4, 5, 6, 7$ or $8$ balls.
Number of ways $= 8$

Answer: (I) $9$ ways; (II) $8$ ways

Jaggu has $3$ red, $4$ black and $5$ blue balls, wherein the balls of each colour are identical. In how many ways
(I) can Jaggu select the balls?
(II) can Jaggu select the balls such that at least one ball is selected?
(III) can Jaggu select the balls such that at least one ball of each colour is selected?

Solution

Case I: Ways of selecting from $3$ red and $4$ black and $5$ blue balls $= (3 + 1) \times (4 + 1) \times (5 + 1) = \bm{120}$

Case II: In the Case I, there is $1$ way where none of the balls are selected. This needs to be removed.

$\therefore$ Number of ways to select $1$ or more of these balls $= 120 - 1 = \bm{119}$

Case III: As selecting none of any of the colours is not allowed, Jaggu can select the $3$ red, $4$ black and $5$ blue balls in $3, 4$ and $5$ ways respectively.

$\therefore$ Number of ways to select $1$ or more of each colour $= 3 \times 4 \times 5 = \bm{60}$

Answer: (I) $120$; (II) $119$; (III) $60$

How many different factors does the number $18000$ have?

Solution

Prime factorising $18000 =$ $2^{4} \times 3^{2} \times 5^{3}$

As there are $4$ identical $2$s, $2$ identical $3$s and $3$ identical $5$s,

Number of factors $= (4 + 1) \times (2 + 1) \times (3 + 1) = 60$

Answer: $60$

In how many ways can $15$ identical balls be placed in $3$ bags of different colours, such that
(I) each bag contains any number of balls?
(II) each bag has at least $1$ ball?
(III) red and blue coloured bags have at least $2$ balls each and black coloured bag has at least $1$ ball?

Solution

Let $a, b$ and $c$ represent the $3$ bags of different colours.

(I) With no other condition, we can apply the formula directly. Let $n$ be the total $15$ and $r$ be the $3$ variables.
$a + b + c = 15$
Number of ways balls can be distributed $=$ $^{15 + 3 - 1}C_{3 - 1} =$ $^{17}C_{2} = 136$

(II) Since there should be at least one ball in each bag, we can begin by placing one ball in each bag. We are now left with $12$ balls.
$a + b + c = 12$
Number of ways balls can be distributed $=$ $^{12 + 3 - 1}C_{3 - 1} =$ $^{14}C_{2} = 91$

(III) Once again, we satisfy the conditions by placing $2, 2$ and $1$ balls in red, blue and black bags respectively. Now, we can apply the formula, where $a + b + c = 10$
$^{10 + 3 - 1}C_{3 - 1} =$ $^{12}C_{2} = 66$

Answer: (I) $136$; (II) $91$; (III) $66$

In how many ways can one or more of $15$ identical balls be placed in $3$ bags of different colours?

Solution

The difference in this question (as compared to the earlier ones) is that there is no conditions for all $15$ balls to be placed. Let's create a dummy variable, say D, which holds the leftover/unplaced balls.
$\therefore$ The equation now is $a + b + c + D = 15$

Number of ways balls can be assigned $=$ $^{15 + 4 - 1}C_{4 - 1} =$ $^{18}C_{3} = 816$

As the questions requires us to find the cases where $1$ or more balls are placed, we need to remove the $1$ case where $a, b$ and $c$ are $0$ and the dummy variable $D = 15$.

$\therefore$ Number of ways in which balls can be assigned $= 816 - 1 = \bm{815}$

Answer: $815$

In how many ways can $15$ identical balls be placed in either red, blue or black bags, such that
(I) each bag has an odd number of balls?
(II) the number of balls in the black bag is a multiple of $4$?

Solution

Case I: Odd numbers are of the form $2k + 1$.
$\therefore$ Let the number of balls in the red, blue and black bags be $(2a + 1), (2b + 1)$ and $(2c + 1)$ respectively.
$(2a + 1) + (2b + 1) + (2c + 1) = 15$
⇒ $2(a + b + c) = 12$
⇒ $a + b + c = 6$

$\therefore$ Number of ways $=$ $^{5 + 3 - 1}C_{3 - 1} =$ $^{8}C_{2} =$ $\bm{28}$

Case II: In this question we have to note that there is no requirement for each bag to have at least one ball. So, there can also be $0$ balls in a bag. Let $c$ be the number of balls in the black bag.

Different cases where balls in the black bag is a multiple of $\bm{4:}$

$c$ =	Number of ways
$0$	$a + b = 15$ ⇒ $^{15 + 2 - 1}C_{2 - 1} =$ $^{16}C_{1} = 16$
$4$	$a + b = 11$ ⇒ $^{11 + 2 - 1}C_{2 - 1} =$ $^{12}C_{1} = 12$
$8$	$a + b = 7$ ⇒ $^{7 + 2 - 1}C_{2 - 1} =$ $^{8}C_{1} = 8$
$12$	$a + b = 3$ ⇒ $^{3 + 2 - 1}C_{2 - 1} =$ $^{4}C_{1} = 4$

Total ways of assigning the balls $= 16 + 12 + 8 + 4 = \bm{40}$

Answer: (I) $28$; (II) $40$

In how many ways can $15$ identical balls be placed in either red, blue or black bags, such that none of the bags contain more than $8$ balls?

Solution

This can be solved by subtracting the ways where the condition is breached from the total number of ways.

$a + b + c = 15$
$\therefore$ Total ways $=$ $^{15 + 3 - 1}C_{3 - 1} =$ $^{17}C_{2} = 136$
Breach in condition occurs if a bag has $9$ balls. Let's first place $9$ balls in the red bag, which leaves us with $6$.
$a + b + c = 6$

$\therefore$ Breaches where red has $9$ or more balls $=$ $^{6 + 3 - 1}C_{3 - 1} =$ $^{8}C_{2} = 28$

Likewise, $28$ such breaches can occur in each of blue and black bags.
$\therefore$ Total breaches $= 28 \times 3 = 84$

Total possible cases $= 136 - 84 = 52$

Answer: $52$

What is the number of distinct terms in the expansion of $(a + b + c)^{20}$?
[CAT 2008]

(1) $231$            (2) $253$            (3) $242$            (4) $210$            (5) $228$

Solution

The powers of $a, b$ and $c$ in each term of the expansion will add up to $20$.
Let $x, y$ and $z$ be the respective powers.
$x + y + z = 20$

Number of different terms $= ^{20 + 3 - 1}C_{3 - 1} = ^{22}C_{2} = 231$

Answer: (1) $231$