# How many poker hands have a pair? (two cards in one denomination)

Poker is played with a regular deck of cards which contains 52 cards. In the deck of cards there are 4 suits (♠♡♢♣), each in 13 denominations. A hand of poker consists of 5 cards.

I need help with the following problem.

How many poker hands have a pair? Meaning, 2 cards in one denomination and 3 cards of other values, the other 3 cards can't be of the same value!

Now they solved it like this, using binomial notation $C(n,k) = \binom{n}{k}$:

• Pick 1 of 13 denominations $C(13,1)$
• Pick 2 of 4 colors $C(4,2)$
• Now pick 3 of 12 denominations $C(12,3)$
• Now each of the three values can be combined with 1 of 4 suits $C(4,1)^3 = 4^3$

You get the following $C(13,1) \cdot C(4,2) \cdot C(12,3) \cdot 4^3 = 1098240$ which is correct.

But why don't they write it like this:

• Pick 4 of 13 denominations $C(13,4)$
• Pick 2 of 4 suits for one of the value above $C(4,2)$
• Now each of the three other values can be combined with 1 of 4 suits $C(4,1)^3 = 4^3$.

You get the following $C(13,4) \cdot C(4,2) \cdot 4^3 = 274560$ (Why is this wrong!?)

Note! $274560\cdot 4=1098240$ (if this might help your explanation for me)

• I think your 27460 should be 274560. Then see Andre Nicolas's answer. Sep 21, 2014 at 14:15
• Yes you are right, I'll change that right away. I don't know why I left the 5 out :P Sep 21, 2014 at 14:51

From the $4$ denominations that we have chosen, we must still choose the one in which we will have a pair, or equivalently the three in which we will have singles.
That's because for example $\spadesuit$K, $\heartsuit$K, $\diamondsuit$$2, \spadesuit$$3$, $\clubsuit$J is different from $\spadesuit$$2, \heartsuit$$2$, $\diamondsuit$K, $\spadesuit$$3$, $\clubsuit$J.