Combinatorics; Distributing meat to children Let there be $4$ kids; people represented as $A, B, C,$ and $D,$ respectively. I'm distributing $6$ strips of bacon and $6$ pork slices to these kids. How many ways can I distribute these items if each kid gets exactly $3$ pieces of meat.
There are $6$ pieces of $2$ different types of meat each. I would think to just use $\displaystyle{\binom{12}{3}},$ since there are $12$ pieces of meat in total. There are four kids, so it would just be $4\cdot\displaystyle{\binom{12}{3}}=\boxed{880}.$
Is there something wrong with my strategy?
 A: It would be right if you had $12$ distinguishable items, and you wanted to count the ways to pick a child and $3$ of the $12$ items and give those $3$ items to the chosen child, but that’s a completely different problem. You don’t have $12$ distinguishable items: the bacon strips are indistiguishable, and so are the pork slices. And you aren’t just picking $3$ of them to give to one of the children.
Notice that if you know how many strips of bacon each child gets, you also know how many pork slices each child gets, because each child gets $3$ pieces altogether. Thus, we we’re really being asked to find out in how many ways $6$ strips of bacon can be distributed amongst $4$ children if each child is to get at most $3$ strips. This is a small enough problem that you can do it by brute force. It turns out that there are $5$ different kinds of distributions:


*

*one child receives $3$ strips of bacon and each of the other children receives one strip;

*one child receives $3$ strips, one receives $2$ strips, and one receives $1$ strip;

*two children receive $3$ strips each;

*three children receive $2$ strips each; and

*two children receive $2$ strips each, and the other two children receive $1$ strip each.


Try to count the distributions of each of these five types; then you can simply add those five numbers to get the answer to the original question. I’ll get you started with the third type in that list: there are $\binom42=6$ ways to choose two of the four children to receive $3$ strips each, so there are altogether $6$ distributions of this type.
