# 5 card poker hand double counting cards

This equation calculates the number of ways to get a pair in a 5 card poke hand. How come this equation doesn't need to be divided by two? My rational for dividing by 2 is as follows: The first two operations calculate getting a pair, and the second two operations calculate getting the other 3 cards in a five card hands. But doesn't multiplying these together imply that order matters. So doesn't it inherently double your answer, because it assumes that pair,other,other,other, is different from other,other,other,pair.

$${13 \choose 1}{4 \choose 2}{12 \choose 3}{(4^3)}$$

Think of it as follows: Multiplying the number of different possibilities for a pair A A and for the remaining cards B C D (disregarding order among the three, of course) together, gives you the number of combinations for hands of the form A A B C D, where each letter denotes some card and reorderings of B C D are not counted seperately. This amounts to disregarding order altogether.
If you were interested in the number of possibilities where the placement of the pair mattered, you'd have to multiply your result by the number of different positions that the pair cards can take in your hand (which is $5\choose 2$). Then you'd be counting e.g. A B A C D and B C D A A as seperate possibilities.