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How many 7-card hands can be dealt having exactly 3 of a kind?

I think the answer is $13\times{4\choose 3}\times{48\choose 4} = 10118160$. Can anyone verify?

13 is for the possibilities of the rank of the 3 of a kind, {4\choose 3} is for the selection of 3 cards from that rank, and {48\choose 4} is for the selection of the remaining 4 cards from the 48 cards that are not in the selected rank.

How many 7-card hands can be dealt having exactly 3 of a kind?

I think the answer is $13\times{4\choose 3}\times{48\choose 4} = 10118160$. Can anyone verify?

13 is for the possibilities of the rank of the 3 of a kind, ${4\choose 3}$ is for the selection of 3 cards from that rank, and ${48\choose 4}$ is for the selection of the remaining 4 cards from the 48 cards that are not in the selected rank.

I think the answer is $13\times{4\choose 3}\times{48\choose 4} = 10118160$. Can anyone verify?

How many 7-card hands can be dealt having exactly 3 of a kind?

I think the answer is $13\times{4\choose 3}\times{48\choose 4} = 10118160$. Can anyone verify?

13 is for the possibilities of the rank of the 3 of a kind, {4\choose 3} is for the selection of 3 cards from that rank, and {48\choose 4} is for the selection of the remaining 4 cards from the 48 cards that are not in the selected rank.

I think the answer is $13*{4\choose 3}*{48\choose 4} = 10,118,160$. Can anyone verify?

I think the answer is $13\times{4\choose 3}\times{48\choose 4} = 10118160$. Can anyone verify?