How can overcounting permutations help us count combinations?

"To see how to count these in general, let’s return to the golf tournament example. Suppose that in addition to winning money, the top three finishers of our local tournament will also advance to the regional tournament. This is a great honor, and brings with it far greater additional winning potential than the local money did. Question: how many different possible trios might we send to regional competition? At first glance, this seems just like the “how many prize money allocations” problem from before, except that we’re taking 3 instead of 10. But there is a twist. In the former problem, it mattered who was first vs. second vs. third. Now the order is irrelevant. If you finish in the top three, you advance, period. You don’t “advance more forcefully” for finishing first locally instead of third. It’s not as obvious how to count this, but of course there is a trick. The trick is to count the partial permutations, but then realize how much we overcounted, and then compensate for it accordingly. If we count the partial permutations of 3 out of 45 golfers, we have 45 × 44 × 43 such permutations. One of those partial permutations is: 1. Phil 2. Bubba 3. Tiger. Another one is: 1. Phil 2. Tiger 3. Bubba, and yet another is: 1. Tiger 2. Phil 3. Bubba. Now the important thing to recognize is that in our present problem — counting the possible number of regional-bound golf trios — all three of these different partial permutations represent the same combination. In all three cases, it’s Bubba, Phil, and Tiger who will represent our local golf association in the regional competition. So by counting all three of them as separate partial permutations, we’ve overcounted the combinations. Obviously we want to count Bubba/Phil/Tiger only once. Okay then. How many times did we overcount it when we counted partial permutations? The answer is that we counted this trio once for every way it can be permuted. The three permutations, above, were examples of this, and so are these three: 1. Tiger 2. Bubba 3. Phil; 1. Bubba 2. Tiger 3. Phil; 1. Bubba 2. Phil 3. Tiger. This makes a total of six times that we (redundantly) counted the same combination when we counted the partial permutations. Why 6? Because that’s the value of 3!, of course. There are 3! different ways to arrange Bubba, Phil, and Tiger, since that’s just a straight permutation of three elements. And so we find that every threesome we want to account for, we have counted 6 times. The way to get the correct answer, then, is obviously to correct for this overcounting by dividing by 6: (45 × 44 × 43) / 6 = 14,190 different threesomes."