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Thread: 2 combinatorics questions

  1. #1

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    Fanatic Member bugzpodder's Avatar
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    2 combinatorics questions

    Could someone find a proof other than induction for these two equations?

    1) nC1+2(n-1)*(nC2)+3(n-1)^2*(nC3)+...+n(n-1)^(n-1)*(nCn) = n^n

    where n in N and n >= 2

    2) Sum (from 3 to n) [3Pk*nCk] =(3Pn)*2^(n-3)



    by the way, 3Pk is k!/(k-3)! some ppl like to write kP3
    Last edited by bugzpodder; Jan 27th, 2003 at 11:21 AM.
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    The probability that a random rational number has an even denominator is 1/3 (Salamin and Gosper 1972)? This result is independently verified by me (2002)!

  2. #2
    vbuggy krtxmrtz's Avatar
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    Re: 2 combinatorics questions

    Originally posted by bugzpodder
    nC1+2(n-1)*(nC2)+3(n-1)^2*(nC3)+...+n(n-1)^(n-1)*(nCn) = n^n
    Is this a typo?

    I may be wrong but are you sure the formula should not begin with "1" instead of "n"? i.e.

    C1+2(n-1)*(nC2)+3(n-1)^2*(nC3)+...+n(n-1)^(n-1)*(nCn) = n^n

  3. #3

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    Fanatic Member bugzpodder's Avatar
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    looks fine to me! nCk=n choose k=n!/[(n-k)!k!]
    Massey RuleZ! ^-^__Cheers!__^-^ Massey RuleZ!


    Did you know that...
    The probability that a random rational number has an even denominator is 1/3 (Salamin and Gosper 1972)? This result is independently verified by me (2002)!

  4. #4
    vbuggy krtxmrtz's Avatar
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    Right, I was misled probably because the first C term was not enclosed within parenthesis and after 8 hours work my brain is somewhat overheated.
    Just wanted to make sure before I tried my luck on this one.

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