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?

I am writing this answer in case you also need to know about the second way: (Ignore if you know the details of the second method.)
Since you need to use each letter atleast once you really have choice only over one of the letters. This can be chosen in $4$ ways and they can be permuted in $\dfrac{5!}{2!}$ ways.
Hence the number of words, is, $\dfrac{5!}{2!} \cdot 4 =240$ words. $\blacksquare$
