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How many possible sequences of length 64 and made from the characters 0123456789ABCDEF are there, where each character appears exactly 4 times.

(This is no homework! I am trying to calculate an upper bound for the number of possible games in the game of "score four")

thank you! kind regards Patrick

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The number you are looking for is called a multinomial coefficient. It is equal to $$ {64 \choose 4, 4, \dots, 4} = \frac{64!}{(4!)^{16}}, $$ which is a very large number indeed.

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Answer. $$ N=\binom{64}{4}\binom{60}{4}\cdots\binom{8}{4}\binom{4}{4}=\frac{64!}{(4!)^{16}}\approx 1.047\times 10^{67} $$

So, we place the 1s in $\binom{64}{4}$ ways, next the 2s in $\binom{60}{4}$ ways etc.

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Start by placing the $0$ four times. There are $\binom{64}{4}$ possibilities.

Then place the $1$ four times. There are $\binom{60}{4}$ possibilities.

Et cetera.

This leads to $$\binom{64}{4}\binom{60}{4}\cdots\binom{8}{4}\binom{4}{4}=\frac{64!}{\left(4!\right)^{16}}$$ possibilities.


alternative

There are $64!$ possibilities to put the characters in a sequence of length $64$ if there would be distinction between $4$ characters that are 'of the same sort'. To repair this double counting we must divide by $4!$ and this for any of the $16$ sorts of character. This leads directly to $$\frac{64!}{\left(4!\right)^{16}}$$

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