Question:medium

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

Updated On: Jan 15, 2026
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Correct Answer: 84

Solution and Explanation

It is required to distribute 20 identical balloons among 4 children. Each child must receive an even number of balloons, and at least one balloon. The number of balloons for the ith child is represented as \(2x_i\), where \(x_i\) is a non-negative integer. The equation governing this distribution is \(2x_1 + 2x_2 + 2x_3 + 2x_4 = 20\). Dividing by 2 yields \(x_1 + x_2 + x_3 + x_4 = 10\). The constraint that each child receives at least 1 balloon implies \(x_i \ge 1\). By substituting \(y_i = x_i - 1\), where \(y_i \ge 0\), the equation transforms to \((y_1+1) + (y_2+1) + (y_3+1) + (y_4+1) = 10\), which simplifies to \(y_1 + y_2 + y_3 + y_4 = 6\). The number of non-negative integer solutions to this equation is determined using the stars and bars method. With \(n=6\) (the sum) and \(k=4\) (the number of variables), the number of solutions is given by \(\binom{n+k-1}{k-1} = \binom{6+4-1}{4-1} = \binom{9}{3} = 84\). Therefore, there are 84 ways to distribute the balloons.
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