If 9 red balloons and 6 blue balloons are to be distributed to 4 children, how many distributions are possible if every child must receive a balloon of each color?

To find the number of possible distributions, we need to consider both the red and blue balloons separately.

First, let's determine how many ways we can distribute the 9 red balloons among the 4 children. This can be solved using stars and bars method (also known as sticks and stones method or balls and urns method).

We have 9 red balloons, and we need to distribute them among 4 children, with each child receiving at least 1 balloon of each color. This means we can imagine the distribution as placing the balloons in 4 boxes (one for each child) and ensuring that each box has at least 1 balloon.

Using stars and bars method, we consider the 9 balloons as stars and we use 3 bars (dividers) to separate the balloons into 4 different boxes. The bars divide the balloons into groups for each child.

For example, if we have 4 red balloons, we can represent the distribution as:

* | * * | * * * |

The first group represents the first child, the second group represents the second child, and so on.

Now, the problem is reduced to counting the number of arrangements of 9 stars and 3 bars. The formula to find the number of arrangements is given by:

C(total objects, dividers) = C(9+3, 3) = C(12, 3) = 220

So, there are 220 possible distributions for the red balloons alone.

Next, let's consider the blue balloons. We have 6 blue balloons to distribute among the 4 children following the same condition of each child receiving at least 1 balloon of each color.

Using the same approach with stars and bars, we have:

C(6+3, 3) = C(9, 3) = 84

There are 84 possible distributions for the blue balloons alone.

To find the total number of possible distributions for both red and blue balloons, we multiply the number of distributions for each color: 220 * 84 = 18,480.

Therefore, there are 18,480 possible distributions if each child must receive a balloon of each color.