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.

For  future  reference  here  is  a  verification  of  the  result  by
N.F. Taussig, which  I upvoted. This is the case  where it is possible
for a  child not  to receive any  presents. The  combinatorial species
here is
$$\mathfrak{S}_{=4}(\mathfrak{P}_{\ne 6}(\mathcal{Z})).$$
We get the generating function
$$G(z) = \left(\exp(z)-\frac{z^6}{6!}\right)^4
\\ = \exp(4z) - 4\frac{\exp(3z)z^6}{6!}
+ 6\frac{\exp(2z)z^{12}}{6!^2}
- 4\frac{\exp(z)z^{18}}{6!^3} + \frac{z^{24}}{6!^4}.$$
Extracting coefficients we find
$$20! [z^{20}] G(z) =
20! \left(\frac{4^{20}}{20!} - 4 \frac{3^{14}}{14! \times 6!}
+ 6 \frac{2^8}{8!\times 6!^2} - 4\frac{1^2}{2!\times 6!^3}\right).$$
This evaluates to
$$\bbox[5px,border:2px solid #00A000]{523708416736.}$$
