Number of 5 letter words over a 4 letter group using each letter at least once Given the set $\{a,b,c,d\}$ how many 5 letter words can be formed such that each letter is used at least once?
I tried solving this using inclusion - exclusion but got a ridiculous result:
$4^5 - \binom{4}{1}\cdot 3^5 + \binom{4}{2}\cdot 2^5 - \binom{4}{3}\cdot 1^5 = 2341$
It seems that the correct answer is:
$\frac{5!}{2!}\cdot 4 = 240$
Specifically, the sum of the number of permutations of aabcd, abbcd, abccd and abcdd.
I'm not sure where my mistake was in the inclusion - exclusion approach. My universal set was all possible 5 letter words over a set of 4 letters, minus the number of ways to exclude one letter times the number of 5 letter words over a set of 3 letters, and so on.
Where's my mistake?

Your mistake is in the arithmetic. What you think comes out to 2341 really does come out to 240. 
$4^5=1024$, $3^5=243$, $2^5=32$, $1024-(4)(243)+(6)(32)-4=1024-972+192-4=1216-976=240$
