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How many different Strings can I create with the characters of the String SPEISESTAERKE ?

According to that post, I have tried the following. Note, that I use all the given characters at once to create all possible combinations:

$2 \times$S, $1 \times$P, $4 \times$E, $1 \times$I, $1 \times$A, $1 \times$R, $1 \times$K

$\binom {13}{2} \cdot \binom {11}{1} \cdot \binom {10}{4} \cdot \binom {6}{1} \cdot \binom {5}{1} \cdot \binom {4}{1} \cdot \binom {3}{1} \cdot \binom {2}{1} \cdot \binom {1}{1} = 129 729 600$

Unfortunately this is the wrong answer. The correct solution is $432 432 00$.

Question: How can I achieve the correct result?

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    $\begingroup$ Your letter counts are wrong. I see three S's, for example, and a T. $\endgroup$ – Henning Makholm Mar 31 '17 at 9:44
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We can calculate the easiest case first and then tailor the number for the purpose. We pretend the letters are all different; then, since there are 13 letters, we can obtain $13!$ words. But there are replicate ones. The "S" appears three times and the "E" appears four times. So $13!$ is $3! \times 4!$ times the number required! So the required number is $$ \frac{13!}{3!4!}. $$

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alternative answer - with correctly counted letters:

$\binom {13}{3} \cdot \binom {10}{4} \cdot \binom {6}{1} \cdot \binom {5}{1} \cdot \binom {4}{1} \cdot \binom {3}{1} \cdot \binom {2}{1} \cdot \binom {1}{1} = 432 432 00$

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    $\begingroup$ Note that this calculation is known as a [multinomial coefficient])(en.wikipedia.org/wiki/…) and often notated $\binom{13}{3,4,1,1,1,1,1,1}$. $\endgroup$ – Henning Makholm Mar 31 '17 at 12:40

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