Anonymous

# math12 -Combinatorics?

A 5-card hand is dealt from a deck of 52 cards. How many different 5-cards contain a pair of aces plus another pair?

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Let's try to simplify and then build the answer up.

Here's how I reason:

There are 4 aces in the deck, so there are 4 ways I could pick the first one, and after that there are three other aces I could draw to get my second ace. Because I don't care what order they come in, I divide by two for the two possible orderings of two cards. This is "n choose r" or C(n, r) = n! / ((r!)(n-r)!) = 4! / ((4-2)! (2!)) = 24 / (2 * 2) = 6 (You can verify this one on your fingers.)

Now, for a pair of aces and kings, there are essentially also 6 ways I could draw a pair of kings. Since it doesn't matter whether I got these 4 cards as AAKK, AKAK, AKKA, KAKA or whatever, without regard to order, there are 36 ways I can be holding a pair of kings and aces.

But we're not just interested in aces and kings, it's aces and another pair. After you have a pair of aces, there are 12 other ranks you could get a pair of. And since there are 6 ways to hold any particular pair, I think the number of ways you can hold Aces and another pair is: 6 * 6 * 12.

Now, for each combination of those 4 cards, there are 48 other cards that can make up your 5th card (some of which will form a full house, and some of which don't - see below).

So, all told, I think the answer is 6 * 6 * 12 * (52-4) = 20736

If you mean hands with a pair of aces, another pair, and the 5th card strictly not matching either pair, then it's 6 * 6 * 12 * (52-8) = 19008.

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• Anonymous
4 years ago

9*10^3*25^2 = 5,625,000 ------------ recommendations: There are 9 alternatives for the 1st quantity, 10 alternatives for all of the subsequent 3 numbers, and 25 alternatives for all of the two letters.

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