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Combinatrics

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Combinatrics [#permalink] New post 08 Nov 2009, 13:02
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Before I start, I hope this in the right place. I got this question from another website.

Q: 4 baseball players each stand at different corners of a baseball diamond. The sides of the diamond are all of equal length. Two arrangements of baseball players are considered different only when the relative positions of the players differ. How many different ways can the baseball players arrange themselves around the diamond?

a) 4 b) 6 c) 16 d) 24 e) 256

I got this right, thanks to a guess and a sketch but I am sure there is a way to solve it that is quicker.



OA = B

OE = The total number of arrangements is expressed by a factorial, 4! = 4 × 3 × 2 × 1 = 24, because there are 4 choices for the first player's corner, 3 choices for the second one, and so on. We divide this total by 4 to arrive at the number of different arrangements of players: 24 ÷ 4 = 6.

I get the factorial part, 4*3*2*1 = 24 - but can't figure out why we are dividing by 4?

Thanks for any input.
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Re: Combinatrics [#permalink] New post 08 Nov 2009, 14:50
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ajthomas wrote:
Before I start, I hope this in the right place. I got this question from another website.

Q: 4 baseball players each stand at different corners of a baseball diamond. The sides of the diamond are all of equal length. Two arrangements of baseball players are considered different only when the relative positions of the players differ. How many different ways can the baseball players arrange themselves around the diamond?

a) 4 b) 6 c) 16 d) 24 e) 256

I got this right, thanks to a guess and a sketch but I am sure there is a way to solve it that is quicker.



OA = B

OE = The total number of arrangements is expressed by a factorial, 4! = 4 × 3 × 2 × 1 = 24, because there are 4 choices for the first player's corner, 3 choices for the second one, and so on. We divide this total by 4 to arrive at the number of different arrangements of players: 24 ÷ 4 = 6.

I get the factorial part, 4*3*2*1 = 24 - but can't figure out why we are dividing by 4?

Thanks for any input.


I don't know why there is such a solution in OE. For me it's circular permutation for 4 which is (4-1)!=3!=6.

The number of circular permutations of n different objects is (n-1)!.

Each circular permutation corresponds to n linear permutations depending on where we start. Since there are exactly n! linear permutations, there are exactly n!/n permutations. Hence, the number of circular permutations is the same as (n-1)!.

When things are arranged in places along a line with first and last place, they form a linear permutation. When things are arranged in places along a closed curve or a circle, in which any place may be regarded as the first or last place, they form a circular permutation.

The permutation in a row or along a line has a beginning and an end, but there is nothing like beginning or end or first and last in a circular permutation. In circular permutations, we consider one of the objects as fixed and the remaining objects are arranged as in linear permutation.

Thus, the number of permutations of 4 objects in a row = 4!, where as the number of circular permutations of 4 objects is (4-1)! = 3!.
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Re: Combinatrics   [#permalink] 08 Nov 2009, 14:50
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