One way to think about it, and these problems in general, is to start by counting the possibilities 'naively'. It seems logical that there should be 5! ways to arrange 5 people around a round table, so start with 5!. Then, account for any special circumstances by figuring out whether you actually counted any of the possibilities more than once. In this case, you actually counted each separate possibility five times by using 5!. For instance, you counted these five arrangements as being different, but they're actually the same (since they're just rotations around the table):

(A B C D E)

(B C D E A)

(C D E A B)

(D E A B C)

(E A B C D)

In order to correct for the overcounting, you'll divide by 5. 5!/5 = 4!, or 24.

This works for a wide range of counting problems. Suppose you wanted to know how many ways a class of eight people could be split into two groups of four. Naively, there are 8*7*6*5 ways to select the first group of four (after which the second group is determined). But you've overcounted by doing that, since you actually counted these groups as being different:

A B C D

B C D A

B A C D

... etc.

That is, you counted each different group 4! times. So, the actual answer is (8*7*6*5)/4!, which is equivalent to what you'd get from the combinatorics formula.

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Chelsey Cooley | Manhattan Prep Instructor | Seattle and Online

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