From a 52 card deck, how many 4 card hands are there that
(a) Do not contain a pair? (That is, the ranks are all different)
(b) Have exactly 2 pairs?
Any help would be great! Thanks!
Well I'm pretty sure the answer to part a is 52*48*44*40/4! because you can pick the first card in 52 ways, then if the second isnt a pair you have to avoid 3 cards so its 48, and so on. And then divide by 4! because order doesn't matter.
The second part I'm having more trouble with. I think it is something like 52*3*48*3/4! but I don't think that's exactly right.
Hello, jlt1209!
Your answer is correct!From a 52 card deck, how many 4-card hands are there that
(a) Do not contain a pair? (That is, the ranks are all different)
First, select the ranks of the two pairs.(b) Have exactly 2 pairs?
. . There are: .$\displaystyle {13\choose2} \:=\:78$ choices.
Then select the two pairs.
. . For the first pair, there are: .$\displaystyle {4\choose2} \:=\:6$ ways.
. . For the second pair, there are: .$\displaystyle {4\choose2} \:=\:6$ ways.
Therefore, there are: .$\displaystyle 78\cdot6\cdot6 \:=\:2808 $ posible two-pair hands.
Why can it not be the OP's answer of (52*3)(48*3) though?
I don't see why you cannot first a pick a card out of the 52, then have only 3 cards to pick from to make a pair. And then, out of the 48 remaining, pick another one out of 3 to make the second pair.
It's confusing to me why this is wrong.