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.

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)
$$
\text{number of ways}=4^{20}-\binom{4}{1}\binom{20}{6}3^{14}+\binom{4}{2}\binom{20}{6}\binom{14}{6}2^{8}-\binom{4}{3}\binom{20}{6}\binom{14}{6}\binom{8}{6}1^{2}+0.
$$
