CAT 2025 Lesson : Permutations & Combinations - Identical Groups

In how many ways can $5$ identical balls be placed in $3$ identical bags?

Solution

In these type of questions we have to follow a pattern to find the different cases.

	Bag $1$	Bag $2$	Bag $3$
Case $1$	Highest no. of balls = $5$	Remaining = $0$	Remaining = $0$
Case $2$	2$^{\mathrm{nd}}$ highest no. of balls = $4$	Remaining = $1$	Remaining = $0$
Case $3$	3$^{\mathrm{rd}}$ highest no. of balls = $3$	Remaining = $2$	Remaining = $0$
Case $4$	3$^{\mathrm{rd}}$ highest - case II = $3$	Remaining = $1$	Remaining = $1$
Case $5$	4$^{\mathrm{th}}$ highest no. Of balls = $2$	Remaining = $2$	Remaining = $1$

Therefore, the $\bm{5}$ different ways are $(5,0,0)$, $(4,1,0)$, $(3,2,0)$, $(3,1,1)$, $(2,2,1)$.
Since these are identical bags, cases like $(3,0,2)$, $(2,1,3)$, etc., will be eliminated because of double counting.

Always maintain a structure so that we would not miss any of the required cases.

Answer: $5$

In how many ways can 5 balls of different colours be placed in $3$ identical bags?

Solution

Note: The way to figure out the required number of cases is given in Example 54.

There are $5$ different cases: $(5,0,0)$, $(4,1,0)$, $(3,2,0)$, $(3,1,1)$, $(2,2,1)$

As these are identical bags with distinct balls, if two or more bags (say $r$ bags) have the same number of balls (other than zero), the repetition while counting should be eliminated by dividing by $r!$.

Sr. No.	Cases	Total ways
1	(5, 0, 0)	$^{5}C_{5} =$ 1
2	(4, 1, 0)	$^{5}C_{4}$ $\times$ $^{1}C_{1} =$ 5
3	(3, 2, 0)	$^{5}C_{3} \times$ $^{2}C_{2} =$ 10
4	(3, 1, 1)	$\dfrac{^{5}C_{3} \times ^{2}C_{1} \times ^{1}C_{1}}{2!} =$ 10
5	(2, 2, 1)	$\dfrac{^{5}C_{2} \times ^{3}C_{2} \times ^{1}C_{1}}{2!} =$ 15

Total number of ways $= 1 + 5 + 10 + 10 + 15 = \bm{41}$ ways

Answer: $41$