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Thread: Find n such that (related to combinations)...

  1. #1
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    Exclamation Find n such that (related to combinations)...

    nC2 = nC7.

    n! / 2!(n - 2)! = n! / 7!(n - 7)!

    You can divide n! out from both sides.

    However, I reached...

    1 / 2! = (n - 2)(n - 3)(n - 4)(n - 5)(n - 6) / 7!

    Something seems off. Too many terms are involved for my liking.
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  2. #2
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    Re: Find n such that (related to combinations)...

    Remember that the binomial coefficients, $^nC_i$, are symmetric- $^3C_i$= 1, 3, 3, 1 and $^4C_i$= 1, 4, 6, 4, 1, etc.

    In order that $^nC_j= ^nC_k$, we must have j= n- k.
    Last edited by HallsofIvy; Feb 21st 2018 at 02:39 AM.
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