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: Convert Even Numbers to Natural Numbers

Since only even numbers are allowed, we can write the number of balloons each child receives as:

\[ 2a,\quad 2b,\quad 2c,\quad 2d \quad \text{where } a, b, c, d \in \mathbb{N} \]

Then the total becomes: \[ 2a + 2b + 2c + 2d = 20 \Rightarrow a + b + c + d = 10 \]

So the original problem is transformed into: How many ways can 4 positive integers sum up to 10?

Step 2: Use Combinatorics for Positive Integer Partitions

The number of ways to partition a number \( n \) into \( r \) positive integers is given by:

\[ \text{Number of solutions} = \binom{n - 1}{r - 1} \]

In our case: \[ n = 10, \quad r = 4 \Rightarrow \binom{10 - 1}{4 - 1} = \binom{9}{3} \]

Calculating: \[ \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]

 Final Answer

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

 Summary

• We transformed the problem from even numbers to natural numbers.
• Used the formula for partitioning a number into positive integers.
• The key was: \( a + b + c + d = 10 \) with \( a, b, c, d \geq 1 \)
• Applied: \( \binom{n - 1}{r - 1} \Rightarrow \binom{9}{3} = 84 \)