The word's length with given alphabet

Question

Consider an alphabet $A = \{a, b, c, d\}$. How many words of length 8 can be created using the alphabet $A$, given the fact that it should contain 3 'a' and 2 'b'?

Answer 1

You need to pick $3$ positions for the 'a' : $8 \choose 3$, then $2$ positions for the 'b': $5 \choose 2$. Now you are left with the number of words of length $3$ on an alphabet of size $2$. You have $2^3$ words.

Finally you have ${8 \choose 3} \times {5 \choose 2} \times 2^3=56 \times 10 \times 8=4480$ words.

Answer 2

There is a general formula in combinatorics: if you have $n$ objects of which have $k_i$ are of type $i$, for $1\leq i \leq m$, the number of words you can make is $\frac{n!}{k_1!k_2!\cdots k_m!}$. That's because you first arrange the $n$ objects in one of the $n!$ permutations, then identify all such permutations that are obtained by simply permuting the identical objects of type $1$ (there are $k_1$ of those, so we need to divide by $k_1!$), and then do the same for the other types.