Question: Deck of 52 cards
13 ranks: 2, 3, . . . , 9, 10, J, Q, K, A
4 suits: ♥, ♠, ♦, ♣,
Poker hands
Consists of 5 cards
A one-pair hand consists of two cards having one rank and the remaining three cards having three other ranks
Example: {2♥, 2♠, 5♥, 8♣,K♦}
Question: How many different 5 card hands have exactly one pair?
Answer: $13C1×4C2×12C3×4^3$
My Confusion: I previously asked a question that was too broad. I deleted the post and thought about exactly where my confusion lies.
1st point of confusion: When I thought about why it's $12C3$ and not $12P3$ I thought about changing the order of the ranks in 5♥, 8♣, K♦ to 8♥, 5♣, K♦. We get a different set of cards if we do this so why aren't we using permutations?
2nd point of confusion: When I visualize $4^3$ I imagine the suits in 5♥, 8♣,K♦ changing like a lottery machine. However, the 5, 8, & K are in a specific order here, namely 5, 8, then K. The $4^3$ is supposed to be per combination not per permutation. So how should I visualize $4^3$?
Best Answer
For me it's easier to get the following answer: $13C1\times 4C2\times (12P3\times 4^3)/P3$. First, we choose the rank of a pair, then two suits of this pair, then three different ordered ranks of the remaining three cards, then their suits. In order to get a set (not a sequence), i.e. to make it independent on the order, we have to divide the product by $P3=3!$, since we counted each hand six times. Observe that this answer is the same as yours.
These two confusion cancel out. Here we don't use permutations, since we want to get just a set of cards and not a sequence. We choose just three ranks out of twelfe. It's important that these three ranks are different, so they are distinguishable and, to make thinking easier, we can put them in ascending order. Then to choose the suits we have three different cards in a fixed order and we have to choose suits for them. We can do it in $4^3$ ways.