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

Step 1: Representing the Distribution

Let the number of balloons each of the 4 children receives be \( 2a, 2b, 2c, 2d \), where \( a, b, c, d \) are positive integers (since each child must receive a non-zero even number of balloons). The total number of balloons is 20.

Thus, we have the equation: \[ 2a + 2b + 2c + 2d = 20 \]

Dividing by 2 simplifies this to: \[ a + b + c + d = 10 \]

The problem is now equivalent to finding the number of positive integer solutions to this equation.

Step 2: Applying Stars and Bars

This is a classic combinatorial problem that can be solved using the "stars and bars" method. The number of ways to distribute \( n \) identical items into \( r \) distinct bins, where each bin must have at least one item, is given by the formula:

\[ \binom{n - 1}{r - 1} \]

In our case, \( n = 10 \) (the total sum) and \( r = 4 \) (the number of variables/children). Applying the formula:

\[ \binom{10 - 1}{4 - 1} = \binom{9}{3} \]

Calculating the binomial coefficient:

\[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84 \]

Final Answer

There are: \[ \boxed{84} \] distinct ways to distribute 20 identical balloons among 4 children such that each child receives a non-zero even number of balloons.

 Summary

• Each child's balloons were represented as \( 2k \), converting the problem to finding positive integer solutions.
• The simplified equation was \( a + b + c + d = 10 \).
• The "stars and bars" combinatorial formula \( \binom{n-1}{r-1} \) was used.
• The calculation yielded \( \binom{9}{3} = 84 \).