math

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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  1. 4*3*2*1

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  2. 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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  3. is n !

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

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  4. 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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  5. 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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  6. Well, that was fun, have not done that for a while.

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  7. :-)

  8. 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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