<s>[INST] Inclusion exclusion distribution problem Prompt: How many ways are there to give 20 different presents to 4 different children, so that no child gets exactly 6 presents. All presents are different.
Here's what I tried doing 
Since there are 20 presents to be distributed among 4 children, 
$$C_1 + C_2 + C_3 + C_4 = 20$$
By bars and stars, 
$$\binom{ 20 + 4-1 }{ 4-1 } = \binom{ 23 }{ 3 } = 1771$$
$C_2 + C_3 + C_4 = 14 $ (after giving 6 presents to $C_1$)
Assuming $C_1$ gets 6 present = $\binom{ 14 + 3 - 1 }{ 3 - 1 } =  \binom{ 16 }{ 2 } = 120$
Assuming $C_2$ gets 6 presents = $\binom{ 14 + 3 - 1 }{ 3 - 1 } =  \binom{ 16 }{ 2 } = 120$
Assuming $C_3$ gets 6 presents = $\binom{ 14 + 3 - 1 }{ 3 - 1 } =  \binom{ 16 }{ 2 } = 120$
Assuming $C_4$ gets 6 presents = $\binom{ 14 + 3 - 1 }{ 3 - 1 } =  \binom{ 16 }{ 2 } = 120$
No one gets exactly 6 presents = Total - Everyone gets 6 presents
= $\binom{ 23 }{ 3 } - \binom{ 4 }{ 1 }\binom{ 16 }{ 2 } = 1771 - 480 = 1291$
I am new to combinatorics and not sure if I'm on the right path any help would be appreciated.
 [/INST] The presents are different so the stars and bars formula doesn't work here. For example, the number of ways to distribute $20$ different presents to $4$ children is $4^{20}$ since each present can go to $4$ possible children (think of stamping the name of the child on each present). 
Let $U$ be the set of all distributions of the $20$ different presents without constraints and let $A_i$ be the set of distributions in which child $i$ has exactly 6 presents. Then by PIE, the desired quantity equals
$$
|\bar{A}_1\cap\bar{ A}_2\cap \bar{A}_3\cap \bar{A}_4|=|U|-S_1+S_2-S_3+S_4\tag{1}
$$
where
$$
S_k=\sum_{1\leq i_1<\dotsb<i_k\leq 4}|A_{i_1}\cap\dotsb A_{i_k}|.\tag{2}
$$
From the above discussion $|U|=4^{20}$ and clearly, $$S_4=|A_1\cap\dotsb \cap A_4|=0.\tag{3}$$
To find $S_1$ note that $|A_i|=\binom{20}{6}3^{14}$ for all $i$ since we choose which six of the $20$ presents child $i$ gets and we distribute the remaining $14$ presents among the remaining $3$ children. Hence
$$
S_1=\binom{4}{1}\times\binom{20}{6}3^{14}\tag{4}.
$$
To find $S_2$ note that for $i\neq j$, we have that $|A_i\cap A_j|=\binom{20}{6}\binom{14}{6}2^{8}$ since we distribute $6$ presents to each of $i$ and $j$ and then the remaining $8$ are distributed without constraints. Hence
$$
S_2=\binom{4}{2}\times\binom{20}{6}\binom{14}{6}2^{8}.\tag{5}
$$
Finally, using similar reasoning as above
$$
S_3=\binom{4}{3}\times\binom{20}{6}\binom{14}{6}\binom{8}{6}1^{2}.\tag{6}
$$
Thus, by (1)
$$
\text{number of ways}=4^{20}-\binom{4}{1}\binom{20}{6}3^{14}+\binom{4}{2}\binom{20}{6}\binom{14}{6}2^{8}-\binom{4}{3}\binom{20}{6}\binom{14}{6}\binom{8}{6}1^{2}+0.
$$
</s>