text: # 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.
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