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  1. #1
    MHF Contributor
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    show this

    Can someone pls show that

    nCr = n(n-1)(n-2) ... (n-r+1)
    r(r-1)*...*2*1
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by mathaddict View Post
    Can someone pls show that

    nCr = n(n-1)(n-2) ... (n-r+1)
    r(r-1)*...*2*1
    _nC_r = \frac{n!}{r!(n-r)!} = \frac{n(n-1)..(n-r+1)(n-r)..(1)}{r!(n-r)!} =\frac{n(n-1)..(n-r+1)r!}{r!(n-r)!}=\frac{n(n-1)..(n-r+1)}{r!}

    or do you mean something else?

    Choosing r from n distinct elements, the first may be choosen in n ways the second in (n-1) down to the r-the in (n-r+1) ways, so there are n(n-1)..(n-r+1) ways of doing the selection, but as each combination appears in these choices in ever permutation we are counting each combination r! times so the number of combinations is:

    \frac{n(n-1)..(n-r+1)}{r!}

    RonL
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