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CAT 2022 Slot 1 — QA Question 7

Letters and Letter BoxesEasy

The number of ways of distributing 20 identical balloons among 4 children such that each child gets some balloons but no child gets an odd number of balloons, is

Answer & solution

Answer: 84

Solution

Easy

"No child gets an odd number" means every share is even, so write each share as twice a positive integer. Halving turns it into a standard "positive integer solutions" stars-and-bars count.

1

Force even, positive shares. Let the four children get 2a,2b,2c,2d2a,2b,2c,2d with a,b,c,d≥1a,b,c,d\ge1:

2a+2b+2c+2d=20 ⇒ a+b+c+d=102a+2b+2c+2d=20\ \Rightarrow\ a+b+c+d=10
2

Count positive integer solutions of a+b+c+d=10a+b+c+d=10 by stars and bars (each ≥1\ge1):

(10−14−1)=(93)=9⋅8⋅73⋅2⋅1=84\binom{10-1}{4-1}=\binom{9}{3}=\frac{9\cdot8\cdot7}{3\cdot2\cdot1}=84
84 ways\mathbf{84}\ \text{ways}
CAT 2022 Slot 1 QA Q7: The number of ways of distributing 20 identical balloons among 4 children such that each child gets some ballo — Solution | TheCATExam