Distribution of balloons, pencils, and erasers

CAT 2021 Slot 2 · QA · Medium · Combinatorics

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The number of ways of distributing 15 identical balloons, 6 identical pencils and 3 identical erasers among 3 children, such that each child gets at least four balloons and one pencil, is

Answer

1000

Explanation

First, give each of the 3 children 4 balloons and 1 pencil.

  • Remaining balloons = 15−3×4=315 - 3 \times 4 = 3
  • Remaining pencils = 6−3×1=36 - 3 \times 1 = 3
  • Remaining erasers = 3

Number of ways to distribute NN identical items among 3 children is given by (N+3−13−1)=(N+22)\binom{N + 3 - 1}{3 - 1} = \binom{N + 2}{2}.

  • Ways for 3 balloons: (3+22)=(52)=10\binom{3+2}{2} = \binom{5}{2} = 10
  • Ways for 3 pencils: (3+22)=(52)=10\binom{3+2}{2} = \binom{5}{2} = 10
  • Ways for 3 erasers: (3+22)=(52)=10\binom{3+2}{2} = \binom{5}{2} = 10

Total ways = 10×10×10=100010 \times 10 \times 10 = 1000.

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