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

Answer:

123,552  ways

Step-by-step explanation:

Solution:-

- A deck of cards has a total of 52 cards. There are 4 suits/ranks in total: Hearts, Spades, Diamonds, and Clubs. There are 13 cards in each suit/rank. So, 4 suits/ranks with 13 cards in each = 4 x 13 = 52 cards ( complete deck ).

- Each suit/rank is divided into similar category of cards. These categories are:

   Numbered cards ( From "A or 1 " to "10" ) = 10 cards

   Jack                                                             = 1 card

   Queen                                                         = 1 card

   King                                                             = 1 card

==================================================

Total                                                                = 13 ordinals

- We are given 5 cards. The 5 cards in such a manner:

2 cards from one rank/suit

- Since any particular rank or suit ( Hearts, Diamonds, Clubs, or Spades) has 13 cards. We are first going to "select" 2 cards from 13 ordinals of a rank. We will use combinations " C " to determine the number of ways:

                           $13 C 2 = \frac{13!}{2!*11!} = 78 ways$

Now we need to do is pick 2 cards from each suit for each ordinal ( Total of 4 suits ) for the first pair. We can determine it:

                             $4 C 2 = \frac{4!}{2!*2!} = 6 ways$

Similarly, for the second pair we pick 2 cards from each suit ( Total of 4 suits ) for each ordinal.

   

                            $4 C 2 = \frac{4!}{2!*2!} = 6 ways$

- We are done with 2 cards for 2 different ordinals. To select the last card in the 5 card-hand. We are left with 11 ordinals in each suit/rank. From which we choose 1 card.

                            $11 C 1 = \frac{11!}{1!*10!} = 11 ways$

- And from 4 ordinals we need one rank/suit:

                             $4 C 1 = \frac{4!}{1!*3!} = 4 ways$

- All the events listed above are dependent events. So the total number of ways to hand out the 5 cards such that there are 2 pairs belonging from 2 ranks and last card from 3rd rank. We have:

                = 13C2 * 4C2 * 4C2 * 11C1 * 4C1

                = 78 * 6 * 6 * 11 * 4              

                = 123,552  ways