In how many ways can we distribute $12$ identical apples and $7$ distinguishable cookies among 5 children..?
My answer is $\binom{12+5-1}{12} \cdot 5^7$
Is it correct?
Explanation: Since there is no restriction to distribute the apples, I lined up the 12 apples and split them into 5 groups using the stars method and for the second one, It is only like assigning a cookie to each child.
What if it says, $7$ distinguishable cookies among $5$ children so that everyone gets at least one?
My answer is: $\binom{7}{5} * 5! * 5^2$ We give the kids $5$ cookies and they can be arranged in $5!$ ways, then we normally distribute the remaining cookies.
Am I missing anything? Are there any notes or something you would like to add up or advice?
Your solution of first question is perfectly correct. You didn't miss anything.
For the second question:
Let $A_i$ be the set of all distributions in which $i-$th children don't get a cookie.
Then $|A_i| = 4^7$ for each $i$.
$|A_i\cap A_j|=3^7$ for each $i\ne j$
$|A_i\cap A_j\cap A_k|=2^7$ for each $i\ne j\ne k\ne i$
$|A_i\cap A_j\cap A_k\cap A_l|=1$ for all pairvise different $i,j,k,l$
Now we are interested in $$\Big| \bigcap_{i=1}^n \overline A_i\Big| = \Big|\; \overline{\bigcup_{i=1}^n A_i}\;\Big| = 5^7-\Big|\; \bigcup_{i=1}^n A_i\;\Big| $$We have by PIE: \begin{eqnarray*} \Big|\; \bigcup_{i=1}^n A_i\;\Big| &=& \sum_{i=1}^n \Big|\; A_i\ \Big| - \sum_{i=1}^n \Big|\;A_i\cap A_j\ \Big|+...\\ &=& 5\cdot 4^7-10\cdot 3^7+10\cdot 2^7-5+0=\\ \end{eqnarray*}
So the finaly answer is $5^7-5\cdot 4^7+10\cdot 3^7-5\cdot 2^7+1$.