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?

Another take on @Aryabhata's solution: If we have the set $\{a, a, b, c, d\}$, it can be permuted in $\binom{5}{2 \; 1 \; 1 \; 1}$ ways (multinomial coefficient, lower story records the numbers of repetitions of each symbol), and as the duplicated letter can be selected in 4 ways the result is:
$$
\binom{5}{2 \; 1 \; 1 \; 1} \cdot 4 = \frac{5!}{2! 1! 1! 1!} \cdot 4 = \frac{5!}{2!} \cdot 4 = 240
$$
