Question

You are dealt $5$ cards from a standard deck of 52 cards. How many ways can you be dealt the $5$ 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.)

Answer 1

First we decide what ranks we will use. Then we will pick suits for all of the cards.

We choose the two paired ranks in ${13\choose 2}=78$ ways and the remaining rank in ${11\choose 1}=11$ ways. Then we choose the suits for these cards in ${4\choose 2}{4\choose 2}{4\choose 1}=144$ ways. This gives a total of $78\cdot 11\cdot 144 = \boxed{123552}$ two pair hands.

Answer 2

247,104

Explanation:

Solution

Given that:

The total number of different 5 card combinations is 2,598,960. which calculated as follows:

52!/(47! * 5!) or 52 * 51 * 50 * 49 * 48/( 5 * 4 * 3 * 2)

The number of two pairs in those combinations of 5 cards is 78 * 72 * 44 which arrives at 247,104.

For the first pair you have the choice of 13 ranks, and six ways to make a pair in that rank. for the second pair you have your choice of 12 ranks, and 6 ways to make a pair in that rank.

Now, after putting together the 2 pair, there are 44 cards left to choose from that don’t turn that 2 pair into a full house.