CAT 2022 Slot 1 QA Q2 Permutations & Probability

Question 2 of the Quantitative Ability (QA) section, CAT 2022 Slot 1, with CAT's official answer and a worked solution.

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

TITA: type in the answer. There are no options and no negative marking.

Answer: 84 Permutations & Probability TITA

Solution

  1. Each child gets a positive even number: write child \(i\)'s share as \(2a_i\) with \(a_i \ge 1\).
  2. Then \(2(a_1+a_2+a_3+a_4) = 20\), so \(a_1+a_2+a_3+a_4 = 10\) with each \(a_i \ge 1\).
  3. Positive solutions: \(\binom{10-1}{4-1} = \binom{9}{3} = 84\).

Shortcut. Halve everything first, then apply stars and bars \(\binom{n-1}{r-1}\).

Same concept in other years

See every question on Integer solutions of x + y + z = n with lower or upper bounds.

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