The following problem is from my textbook:
How many distinct 9-letter strings are possible out of the 17-letter string AAABBCCDEFGHIJKLM?
From this question I've learned about the generating function type of approach and after building the relevant polynomial (W|A link)
$$9!\bigg(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}\bigg)\bigg(1+x+\frac{x^2}{2!}\bigg)^2\bigg(1+x\bigg)^{10}$$
the answer seems to be $898264080$ (the coefficient of $x^9$).
Yet it still felt as a black box so I've decided to do it by hand to make sure I understand.
\begin{array}{l|l} \text{Type} & \text{Number} \\ \hline \text{XXXYYZZ}\square\square & \binom{1}{1}\binom{2}{2}\binom{10}{2}\frac{9!}{3!2!2!} \\ \text{XXXYY}\square\square\square\square & \binom{1}{1}\binom{2}{1}\binom{11}{4}\frac{9!}{3!2!} \\ \text{XXX}\square\square\square\square\square\square & \binom{1}{1}\binom{12}{6}\frac{9!}{3!} \\ \text{XXYYZZ}\square\square\square & \binom{3}{3}\binom{10}{3}\frac{9!}{2!2!2!} \\ \text{XXYY}\square\square\square\square\square & \binom{3}{2}\binom{11}{5}\frac{9!}{2!2!} \\ \text{XX}\square\square\square\square\square\square\square & \binom{3}{1}\binom{12}{7}\frac{9!}{2!} \\ \square\square\square\square\square\square\square\square\square & \binom{13}{9}\frac{9!}{1!} \\ \end{array}
So for example the first line means that we have a sting where
- one symbol appears three times (XXX), there is only one candidate
- two symbols appear twice (YY and ZZ), there are only two candidates
- and two more slots are occupied by the symbols which appear once, there are $10$ candidates and we choose $2$
Yet calculating the sum gives another number: $1329365520$ (W|A link).
I've checked it as best as I could, so the mistake seems to be conceptual. I'd greatly appreciate if someone alleviates my misunderstanding by pointing out the error.
