CAT 2025 Lesson : Permutations & Combinations - Identical Groups

6.7 Elements being assigned to Identical groups

These questions can have identical or distinct elements being assigned to identical groups. In both cases, there is no formula. These have to be done manually.

Example 54

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

Example 55

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

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