calendarBack
Quant

/

Modern Maths

/

Permutations & Combinations
ALL MODULES

CAT 2025 Lesson : Permutations & Combinations - Identical Groups

bookmarked

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 55 identical balls be placed in 33 identical bags?

Solution

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

Bag 11 Bag 22 Bag 33
Case 11 Highest no. of balls = 55 Remaining = 00 Remaining = 00
Case 22 2nd^{\mathrm{nd}} highest no. of balls = 44 Remaining = 11 Remaining = 00
Case 33 3rd^{\mathrm{rd}} highest no. of balls = 33 Remaining = 22 Remaining = 00
Case 44 3rd^{\mathrm{rd}} highest - case II = 33 Remaining = 11 Remaining = 11
Case 55 4th^{\mathrm{th}} highest no. Of balls = 22 Remaining = 22 Remaining = 11


Therefore, the 5\bm{5} different ways are (5,0,0)(5,0,0), (4,1,0)(4,1,0), (3,2,0)(3,2,0), (3,1,1)(3,1,1), (2,2,1)(2,2,1).
Since these are identical bags, cases like
(3,0,2)(3,0,2), (2,1,3)(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:
55


Example 55

In how many ways can 5 balls of different colours be placed in 33 identical bags?

Solution

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

There are 55 different cases: (5,0,0)(5,0,0), (4,1,0)(4,1,0), (3,2,0)(3,2,0), (3,1,1)(3,1,1), (2,2,1)(2,2,1)

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

Sr. No. Cases Total ways
1 (5, 0, 0) 5C5=^{5}C_{5} = 1
2 (4, 1, 0) 5C4^{5}C_{4} ×\times 1C1=^{1}C_{1} = 5
3 (3, 2, 0) 5C3×^{5}C_{3} \times 2C2=^{2}C_{2} = 10
4 (3, 1, 1) 5C3×2C1×1C12!= \dfrac{^{5}C_{3} \times ^{2}C_{1} \times ^{1}C_{1}}{2!} = 10
5 (2, 2, 1) 5C2×3C2×1C12!=\dfrac{^{5}C_{2} \times ^{3}C_{2} \times ^{1}C_{1}}{2!} = 15


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

Answer:
4141


Loading...Loading Video....