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!

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- Sep 24th 2008, 02:28 PMjlt1209From 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! - Sep 24th 2008, 03:11 PMPlato
- Sep 24th 2008, 05:34 PMjlt1209
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. - Sep 24th 2008, 08:38 PMSoroban
Hello, jlt1209!

Quote:

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)

Quote:

(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.

- Sep 24th 2008, 08:53 PMjlt1209
Ohh ok that makes sense. Thanks so much! :)

- Sep 25th 2008, 09:48 AMHobbes
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.