# Game of Score Four

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

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.

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.

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 $$\frac{64!}{\left(4!\right)^{16}}$$