How many five card hands can you have if..

if three of the cards have the same face value and the other two have different face values?

So far i have this for a solution

(13C1) * (4C3) * (12C1) * (4C1) * (11C1) * (4C1)

Where xCy is a combination, and 13 or 12 or 11 C 1 is the selection of the face value, and the 4c3 or 1 is the selection of the suit(clubs, hearts, diamonds, or spades).



But i have twice the value that it should be, i'm not sure what i'm doing wrong, if someone can explain this to me intuitively that would be fantastic!

Thanks

ok. take the set (that is, 3 of a kind) first. there are 13 different denominations that set can have. further, there are 4 unique ways to put together a set (there is always one suit void in a set, and there are four different suits to void). so, there are 13 * 4 = 52 different ways to have a set.

when you say, "the other two [cards] have different face values," i'm going to assume that they need to be different than each other AND different than the denomination of the set. so, for the fourth card, you have (12*4) different possibilities - anything but the denomination of the set, four different suits of each denomination - and for the fifth card, you have (11*4) choices remaining. so, 52 * (12*4) * (11*4) = 109824 possibilities total.

answerer here. there are no 2's to remove from the calculation. the only thing i can think to do in order to remove a 2 is to presume there are only 2 ways to have a set of a given denomination, which is not correct. i really think you and i have this one right.

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eeerrrrr.............10