# In poker, a 5-card hand is called two pair if there are two cards of one rank, two cards of another rank, and a fifth card of a third order of the cards doesn't matter (so, example, QQ668 and 68QQ6 are the same). How many 5-card hands are two pair?

$78 \times 6 \times 6 \times 11 \times 4 = 2 , 598 , 960$

#### Explanation:

Let's work this through in bits.

Ok - first to know is that a standard card deck has 52 cards. There are 13 cards (1 - 10, Jack, Queen, King - let's call them Ordinals) in each of four Suits (spades, clubs, hearts, diamonds), which is $4 \times 13 = 52$

Ok, let's now work through the number of 5 card hands that are 2 pairs.

The first two cards will be the first card each of the 2 pairs. We have 13 ordinals to choose from and we need one, so we can express that as:

C_(13,2)=(13!)/((2!)(11!))=78

(I'm using C to mean Combination, the first number being the number of things we can choose from, and the second number being the number being selected). So there are 78 different ways to pick 2 ordinals from a group of 13.

Next thing we need to do is pick 2 cards from each suit for each ordinal. We can write doing it once as:

C_(4,2)=(4!)/((2!)(2!))=6

and we need to do it twice, so let's write the formula again for the other suit:

C_(4,2)=(4!)/((2!)(2!))=6

So now we have 2 cards in one ordinal and 2 in the other ordinal. Two pairs.

Lastly we need to account for the last card in 5 card hand. There are 11 ordinals to pick from and we need only 1:

C_(11,1)=(11!)/((10!)(1!))=11

and from that one ordinal we need 1 suit:

C_(4,1)=(4!)/((1!)(3!))=4

To find the number of 2-pair hands, we multiply it all together:

$78 \times 6 \times 6 \times 11 \times 4 = 2 , 598 , 960$