# math

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annie,betty,Cathy and Danial are going to stand on a row for taking pictures. How many ways can they stand?

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4*3*2*1

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The first person has the cohoice of 4 spots, takes one

The second person has choice of 3 spots, takes one

The third person has choice of two spots takes one

The fourth person is stuck with the last spot, takes it

4*3*2*1 = 24

permutations of n items taken one at a time.

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is n !

n (n-1) (n-2) ....... 1

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By the way the permutations of n items taken r at a time is
n!/(n-r)!

and if it does not matter how the r items are arranged in each sub group, then it is the number of "combinations"
n! / [ r! (n-r)! ]

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Now what if they were to be paired and we care if each is on left or right in the photo?

that is permutations of four people taken 2 at a time

4!/2! = 4*3*2*1/(2*1) = 12 ways

But what if we do not care if they are on the right or the left. (Betty-Cathy the same as Cathy-Betty)
4!/[2!(4-2)!] = 4*3/2 = 6 ways

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Well, that was fun, have not done that for a while.

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:-)

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LOL - well there is much more but it is probably not on Em's list for tonight (Pascal's triangle, binomial theorem). Do not get me started :)

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