In how many ways can I distribute 6 identical cookies and 6 identical candies to 4 children, if each child must receive exactly 3 items? I tried to solve this by making a chain of letters, with 'O' representing cookies and 'A' representing candies, as shown below.
o o o o o o a a a a a a
1 1 1 2 2 2 3 3 3 4 4 4
This would mean that child one gets 3 cookies, as does child 2, and children 3 and 4 get three candies each.
By finding the amount of strings with 6 o's and 6 a's, I get $\frac{12!}{6!6!}$, which gives me 924. Unfortunately, this is incorrect. What am I doing wrong? Or is my whole approach unfounded?
 A: Write $6$ as a sum of $4$ terms (non-negative integers) that do not exceed $3$. The first term gives the number of cookies for child 1, the second gives the number of cookies for child 2, et cetera. If a child gets $t$ cookies then it will get $3-t$ candies.
The number of sums is the number of distributions.
$6=3+3+0+0$ gives $\dfrac{4!}{2!2!}=6$ possibilities
$6=3+2+1+0$ gives $\dfrac{4!}{1!1!1!1!}=24$ possibilities
$6=3+1+1+1$ gives $\dfrac{4!}{1!3!}=4$ possibilities
$6=2+2+2+0$ gives $\dfrac{4!}{3!1!}$=4 possibilities
$6=2+2+1+1$ gives $\dfrac{4!}{2!2!}=6$ possibilities
So there are $44$ sums that suffice.
A: The following approach would be painful with larger numbers, but works fairly well here.
Let us distribute the $6$ cookies so that each child gets no more than $3$. Then the distribution of the candies is determined. 
To count the number of ways to distribute the cookies, first count the number of ways to distribute the cookies with no restriction. By standard techniques (Stars and Bars), there are $\binom{9}{3}$ ways to do this.
Now subtract the bad distributions of cookies, the ones in which some child gets $4$ or more. 
To find this, treat the cases some child gets $6$, some child gets $5$, some child gets $4$ separately. We could use Stars and Bars for each of these, but we will do it in a more simple-minded way. 
Some child gets $6$: The lucky child can be chosen in $\binom{4}{1}$ ways.
Some child gets $5$: The lucky one can be chosen in $\binom{4}{1}$ ways, and for each way the one who gets $1$ cookie can be chosen in $\binom{3}{1}$ ways, for a total of $\binom{4}{1}\binom{3}{1}$.
Some child gets $4$: The child can be chosen in $\binom{4}{1}$ ways. Now for the rest, one child gets $2$ ($\binom{4}{1}$ choices) or $2$ children get $1$ each ($\binom{3}{2}$ ways) for a total of $\binom{4}{1}\binom{3}{1}+\binom{4}{1}\binom{3}{2}$.
Now put things together. 
A: $6$ candies can be divided among $4$ children in ${9\choose 3}=84$ ways ("stars and bars"). Allocations where (at most) one child gets $\geq4$ candies are not admissible and have to be discounted. Such a child can be chosen in $4$ ways, and the remaining $2$ candies can be divided among the $4$ children in ${5\choose 3}=10$ ways. It follows that there are $40$ forbidden allocations, giving a total of $84-40=44$ admissible allocations of the candies, and each of these determines a unique admissible allocation of the chocolates.
