CAT 2022 Slot 1 · Quant · Question 2

Every child must receive an even number of balloons, and at least one balloon, so each share is a positive even number.

Write the shares as 2{a}_{1}, 2{a}_{2}, 2{a}_{3} and 2{a}_{4}, where each {a}_{i} is a positive integer.

Their total is 2\left({a}_{1}+{a}_{2}+{a}_{3}+{a}_{4}\right)=20.

Dividing by 2, {a}_{1}+{a}_{2}+{a}_{3}+{a}_{4}=10 with every {a}_{i}\ge 1.

The number of ways to write 10 as an ordered sum of 4 positive integers is \left( \genfrac{}{}{0pt}{}{10-1}{4-1}\right)=\left( \genfrac{}{}{0pt}{}{9}{3}\right).

Computing, \left( \genfrac{}{}{0pt}{}{9}{3}\right)=\frac{9\times 8\times 7}{3\times 2\times 1}=84.

So there are 84 distributions.

Shortcut. Halving every share is what turns an awkward even-only condition into the standard positive-integer distribution. The gaps-and-bars count \left( \genfrac{}{}{0pt}{}{n-1}{r-1}\right) then applies directly, with no case work.