Yes indeed. (I'll ramble on for a while here, since I already spent three hours tutoring statistics today, and it's sometimes hard to stop.) Although I'm not sure how many people on the board know what your n and r were, and how to figure out that ! thingy. (The ! thingy means multiply the number by all the numbers smaller than it - so 3! = 3*2*1 = 6. Numbers get big fast when you do that.)
The formula is for "how many ways can you choose r of n things". So, if there are four kinds of beer on tap, and you're getting one for your wife as well, how many choices of two of the four are there? 4!/2!(4-2)!, that's how many, which is 4*3*2*1/2*1*2*1, which is 6. It's important to note that this formula assumes you can't choose the same thing twice - wrong for my beer example (sure, you can both have Bud Light if you want), but good in Hold'em where you don't want to consider the same card coming up twice.
The most direct way to get to the original question of how many combinations of AA, KK, AK there are with the formula is to consider that you have 8 cards (four Aces and four Kings), and that you need to know how many ways there are to choose two of them. That's 8 choose 2, or 8!/2!(8-2)!, or 8*7/2 = 28. Voila.
MathBabe
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