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You are dealt  cards from a standard deck of 52 cards.

 

How many ways can you be dealt the  cards so that they contain two cards of one rank, two cards of another rank, and a fifth card of a third rank? We say that such a hand has two pairs. For example, the hand QQ225 has two pairs. (Assume that the order of the cards does not matter.)

 

 

I thought that there would be 13 choices for the first card, 1 for the second, 11 for the third, 1 for the fourth, and 9 for the fifth. Multiply that and get 1287.

However, my answer is showing as wrong. Could someone please explain?

 Jun 26, 2020
 #1
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There are C(4,2)*4*3*4*3*44 = 38016 ways of getting two pairs.

 Jul 19, 2020

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