I know the answer is $C(8,2)$, but I was confused as to why the answer wouldn't be $P(8,2)$? Doesn't the order of the 0s and 1s matter? For example: 10010000 is different than 10100000, so don't we have to consider every possible permutation of the zeros and the two ones?
2026-03-31 07:51:25.1774943485
How many bytes contain exactly two 1s?
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Yes. But, combinations take care of that. There are $8!$ ways to permute the symbols in a byte. If 2 are ones interchanging them ( aka permuting indistinguishable things), doesn't change the combination placement. Same with interchanging the six 0's . So, we have:$$\frac{8!}{2!6!}=\binom{8}{2}$$ ways.