How many ways are there to distribute 8 teachers to 4 schools where each school must get at least 1 teacher? Additional details: the teachers are considered distinct from one another.
So here is what I thought:
1) Choose four teachers to go to each one of the schools: $\binom{8}{4}\cdot4!$
2) For each of those situations, distribute the other 4 teachers to the 4 schools: $4^4$
So total: $\binom{8}{4}\cdot4!\cdot4^4$
However, I am almost 100% sure that I am over-counting but can't quite put my finger on it. Any help would be appreciated.
 A: We can use Inclusion/Exclusion. There are $4^8$ ways to assign the $8$ teachers, with no restrictions. 
We need to remove the bad assignments, where some school(s) get no teacher. 
Let the schools be A, B, C, D. There are $3^8$ ways to assign the teachers, avoiding school $A$. There are also $3^8$ ways to avoid schools B, C, D, for a total of $\binom{4}{1}3^8$.
However, this overcounts the bad assignments. For example, $\binom{4}{1}3^8$ counts twice the assignments that avoid schools A and B. The same is true for all the $\binom{4}{2}$ pairs of schools. So to count the bad assignments, we must subtract $\binom{4}{2}2^8$.  
But we have subtracted too much, for we have subtracted one too many times the assignments that avoid all schools but A, also the ones that avoid all schoola but B, or all but C, or all but D. So we must add back $\binom{4}{3}1^8$. 
The total number of good assignments is therefore 
$$4^8-\binom{4}{1}3^8+\binom{4}{2}2^8-\binom{4}{3}1^8.$$
Remark: The counting is presumably not being done by a School Board, since a Board is likely to consider teachers indistinguishable. 
A: Another way is using exponential generating functions.
The EGF of this problem is:
$\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+
\frac{x^7}{7!}+\frac{x^8}{8!}\right)^4$
to solve the problem we can use 
$(e^x-1)^4$
for that expression. This expands out to
$ 1-4 e^x+6 e^{2 x}-4 e^{3 x}+e^{4 x}$
this can expanded by hand or taken over to Wolfram Alpha. We get:
$x^4+2 x^5+\frac{13 x^6}{6}+\frac{5 x^7}{3}+
\frac{81 x^8}{80}+...+$
The coefficient of x^8 is the one we want and we multiply that by 8!
$\dfrac{81}{80}\cdot 8! = 40824$
That is the number of ways.
