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January 26, 2015

January 26, 2015

Posted by **K.** on Wednesday, January 31, 2007 at 7:33am.

If person A can be in any of four positions, then person B can be in any of three, and so on.

Combination= n! or in this case 4*3*2*1

Suppose you have N people. Then there are N choices for choosing which person will stand to the right of everyone else. Clearly all these N choices for choosing which person will stand on the right will lead to different combinations.

Once you have made your choice for the person who will stand on the right, you have N-1 persons left. You can then choose which of these persons will stand next to the first person you have chosen. For a given choice of the first person all these N-1 choices will lead to different combinations. In fact, if you keep the choice for the first person fixed, the total number of combinations is just what the answer to the original problem would be if N were replaced by N-1.

So, we can conclude that for N persons the number of combinations is N times the number of combinations for N-1 persons. Now for 1 person the number of combinations is, of course, 1. This means that for 2 it is 2*1 = 2. For 3 it is 3*2*1 = 6, for 4 it is 4*6 = 24, etc.

The notation N! (factorial N) is used to denote this quantity:

N! = N*(N-1)*(N-2)*...*1

good post.

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