# From a 52 card deck, how many 4 card hands ...

• September 24th 2008, 02:28 PM
jlt1209
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!
• September 24th 2008, 03:11 PM
Plato
Quote:

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?

• September 24th 2008, 05:34 PM
jlt1209
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.
• September 24th 2008, 08:38 PM
Soroban
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?
First, select the ranks of the two pairs.
. . There are: . ${13\choose2} \:=\:78$ choices.

Then select the two pairs.
. . For the first pair, there are: . ${4\choose2} \:=\:6$ ways.
. . For the second pair, there are: . ${4\choose2} \:=\:6$ ways.

Therefore, there are: . $78\cdot6\cdot6 \:=\:2808$ posible two-pair hands.

• September 24th 2008, 08:53 PM
jlt1209
Ohh ok that makes sense. Thanks so much! :)
• September 25th 2008, 09:48 AM
Hobbes
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.