How many ways can the $12$ presents be distributed among $12$ children if presents are of $2$ different types, $6$ presents of each type? How many ways can the $12$ presents be distributed among $12$ children if presents are of $2$ different types, $6$ presents of each type?
I have tried finding the answer to this question but was unable to. Please can someone help me? 
I tried doing $12!$
 A: If each child gets one present, we simply pick 6 out of 12 children to receive the first type. Answer is $\binom{12}{6}$
If it is possible to receive more than one present, define $a_{i}$ and $b_{i}$ as the amount of first type and second type present that child $i$ receive respectively.
We then have $\sum_{i=1}^{12}{a_{i}}=6$ and $\sum_{i=1}^{12}{b_{i}}=6$. Stars and bars and then multiplication to obtain $\binom{17}{5}\times\binom{17}{5}$
A: If the presents of one type are all identical, we must simply pick out 6 to get the first type and then automatically the remaining children would get the other type. The answer would simply be $C(12,6)$.
If the presents of one type are different in some way, we must first choose 6 from 12 and then find the ways of giving or 'arranging' six presents among six people. That would be:
$C(12,6)$$6!$ 
If a child, can receive more than one present, things become interesting:
First we need to select 6 from 12 as usual,
then, we have the condition that each child can get any number of presents, but the total number must add up to 6, for each case:
let $c_1$, $c_2$,.....,$c_6$ denote the number of presents of any one type received by the children, then the only condition is that
$c_1$ + $c_2$ + .... + $c_6$ = 6 (the total number of presents here must add up to 6)
Its quite a well known result then, that the number of solutions to this equation is given by $C(6 + 5, 5)$ = $C(11, 5)$
do the same for the other type of present, then multiply to get:
$C(11,5)$$C(11,5)$.
Use this to find the answer, if anyone finds any errors please let me know and ill try to fix it :) 
Edit: Ive assumed here that one child shall get just one type of present. If it is possible to give both types to one child, then by a very similar method:
The answer would be: $C(17,5)$$C(17,5)$
