Thread: From a 52 card deck, how many 4 card hands ...

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!

Originally Posted by jlt1209

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?

Why don't you tell us what you know about this problem?
Just giving you a complete solution will not help you understand anything about this subject.

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!

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)

Your answer is correct!

(b) Have exactly 2 pairs?

First, select the ranks of the two pairs.
. . There are: . choices.

Then select the two pairs.
. . For the first pair, there are: . ways.
. . For the second pair, there are: . ways.

Therefore, there are: . posible two-pair hands.

Ohh ok that makes sense. Thanks so much!

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.