View Full Version : Is this math right?

02-13-2003, 04:50 PM
The question is, how many ways can you be dealt a pair of aces and a kicker 8 or lower, from a standard 52 card deck. Trip aces is excluded.

I thought it was done like this. There are 28 cards that can be kickers (2 through 8).

The first card can be 1 of 28, the kicker.
The second must be 1 of 4 aces.
The third must be 1 of the other 3 aces.

28*4*3 = 336.

However, we've overcounted because of the aces. We have to make the order of the aces not count, by dividing the whole thing by 2!

336/2! = 168

My questions are...

1) Is 168 correct?

1a) If so , why don't we have to divide by 3! ( the number of ways 3 cards can be mixed up)?

1b) If not , what is the answer?

Thanks in advance

02-13-2003, 06:15 PM
1) Yes, 168 is correct. Look at it this way, there are 6 ways to get AA, and for each of those there are 28 other cards 2-8 to go with it, so 6*28 = 168.

1a) This kind of thing is the source of potential confusion. Your question is essentially why do we divide by 2 and not 6? The answer is because you already divided by 3 when you said that the the 1st card is the kicker. If you had counted every possible order to start with, then you would have said that the kicker could be the 2nd or 3rd card also, and there would be 3 times more orderings 28*4*3*3. Then you can divide by 3!=6 to get 168.

In many problems you have a choice as to whether to start out counting all possible orderings (permutations) or just combinations that go together, and you just have to know which one you are doing and be consistent to get the answer you want. In this case it is easiest to think in terms of combinations 6*28. Whenever you specify a particular ordering of n things by saying "first we'll pick this one and then we'll pick that one, etc.", there will be n! times more ways if you had instead counted every possible ordering. This should not be confused with the similar sounding but different case where you have a pair like a pair of aces and you say "first we'll pick an ace and then we'll pick another ace". This isn't the same thing because we aren't specifying which ace we are picking first, so we are actually specifying every possible ordering, and you would have to divide by 2! = 2 to get rid of the different orderings.

You said "first I'll pick the kicker, then I'll pick one of the 4 aces, then I'll pick one of the 3 remaining aces". This gets rid of order for the kicker by making it first, but not for the aces, so we divide by 2.

02-14-2003, 11:02 AM