Why am I not correct when calculating the number of ways of selecting a pair when a five-card hand is drawn? Discrete Maths noob here.
I'm trying to calculate amount of possible choices when choosing exactly one pair out of a deck of cards.
I'm thinking that you could first choose any of the 52 cards, and that your next choice must one out of three cards (because you chose a value in your first choice), after that you can choose any out of the 48 remaining cards (subtracting the four varieties of the first card chosen), and then you can choose 1 out of 44, and lastly one out of 40.
$$
{{52}\choose{1}}\cdot{{3}\choose{1}}\cdot{{48}\choose{1}}\cdot{{44}\choose{1}}\cdot{{40}\choose{1}}
$$
But this results in a really wrong answer.
Any suggestions on how to think?
 A: You're trying to calculate the number of choices which give you exactly one pair out of five cards, right?
What you've actually calculated is the number of choices where the first two cards form a pair, and the rest are all different. Since there are actually $\binom 52=10$ different choices for which two of the five cards make up the pair, you have to multiply the answer by $10$. 
(edit: that is, you have to multiply by $10$ if you want the number of ordered choices of five cards which contain exactly one pair. If you just want the number of hands of five cards, unordered, you have to divide the ordered answer by $5!=120$ (the number of ways to order a given hand), so as Wen says you end up with your answer divided by $12$.)
A: Especially Lime has provided you with a nice explanation of why your answer is incorrect.
How can we calculate the number of five-card hands that contain one pair?
Such hands contain two cards of one rank and one card each of three other ranks.  There are $13$ ways to choose the rank from which the pair will be drawn.  There are $\binom{4}{2}$ ways to select two of the four cards from this rank.  There are $\binom{12}{3}$ ways to select the ranks of the other three cards in the hand from the $12$ other ranks in the deck.  There are four ways to select one card from each of these ranks.  Hence, the number of five-card hands that contain a pair is 
$$\binom{13}{1}\binom{4}{2}\binom{12}{3}\binom{4}{1}^3$$
