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:

Question

Accepted Answer

To solve the problem, we use the condition that each child receives an even number of balloons. Let the number of balloons received by the four children be $$2a$$, $$2b$$, $$2c$$, and $$2d$$, where $$a, b, c, d$$ are natural numbers.

1. **Equation Setup:**  
   Since the total number of balloons is 20, we have:  
   $$
   2a + 2b + 2c + 2d = 20
   $$
   Dividing through by 2:  
   $$
   a + b + c + d = 10
   $$

2. **Number of Solutions:**  
   The problem now reduces to finding the number of ways to distribute 10 as a sum of four natural numbers. Using the formula for partitioning $$n$$ into $$k$$ parts ($$n-1 \choose k-1$$):  
   $$
   	ext{Number of solutions} = \binom{10-1}{4-1} = \binom{9}{3}
   $$

3. **Calculation:**  
   $$
   \binom{9}{3} = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84
   $$

Thus, the number of ways to distribute the balloons is **84**.