How many ways can you give 2 cookies and 2 candies to 4 children?

Question

The title basically sums up the whole question. How many ways can you give 2 identical cookies and 2 identical candies to 4 children without restriction?

Answer 1

$$\frac{4!}{2! 2!} = 6$$ There are 6 ways

Answer 2

Let's make up words that will tell us how to hand out the items. We'll use $K$ for cookie, $C$ for candy and a bar $|$ to separate them between the four kids. We'll use 3 bars, and we'll give out to the first kid the stuff that appears before the first bar, to the second kid the stuff that appears between the first bar and the second, and so on.

We hand out the candies first, which is rearranging the word $CC|||$. For example, $C|C||$ means give a candy to the first kid and one to the second kid, and no candy to the third and fourth kids.

We have $\frac{5!}{2!3!}$ ways to do this.

Then we hand out the cookies, which reduces to the same thing, rearranging $KK|||$, so $\frac{5!}{2!3!}$ ways for this task.

The total number of ways is the product of the number of ways for each task, so we have

$(\frac{5!}{2!3!})^2 = 100$ ways of doing it.

Edit: A clarification: we can't do it all at once rearranging $KKCC|||$, as we would be counting many times some configurations, for example $CK|CK||$, $KC|KC||$, $CK|KC||$, etc., which are the same. This would give us the wrong answer of $210$. It's difficult to "cancel out" this kind of repetitions if we do it this way, but easy when we do it in two steps: when we divide by $2!$ we're accounting for the permutations of the $C$ (or $K$) which we are counting separately in the $5!$ part, and dividing by $3!$ gets rid of the repetitions arising from the $|$ switching places.