Question

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 1

Correct Answer: 84

We need to distribute 20 identical balloons among 4 children so that each child receives an even number of balloons and at least one balloon. To achieve this, let's define how many balloons each child receives in terms of even numbers.

Let the number of balloons received by the ith child be denoted as \(2x_i\), where \(x_i\) is a non-negative integer. This ensures each child gets an even number of balloons. Our task is to solve the equation:

\(2x_1 + 2x_2 + 2x_3 + 2x_4 = 20\)

Dividing the entire equation by 2 gives us:

\(x_1 + x_2 + x_3 + x_4 = 10\)

We must find the number of non-negative integer solutions to this equation, ensuring each child receives at least 1 balloon, i.e., \(x_i ≥ 1\). To simplify further, let's substitute \(y_i = x_i - 1\) which means \(y_i ≥ 0\). This transforms the equation to:

\((y_1+1) + (y_2+1) + (y_3+1) + (y_4+1) = 10\)
\(y_1 + y_2 + y_3 + y_4 = 6\)

We're looking for the number of combinations of non-negative integer solutions to this equation, which can be calculated using the stars and bars method. The formula for the number of solutions is given by:

\(\binom{n+k-1}{k-1}\)

Here, \(n=6\) (total sum we want) and \(k=4\) (number of children), so the number of solutions is:

\(\binom{6+4-1}{4-1} = \binom{9}{3} = 84\)

Thus, the number of ways to distribute the balloons is 84, which is within the expected range [84, 84].