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Permutation proof

  1. Oct 24, 2015 #1
    1. The problem statement, all variables and given/known data
    Hello
    I have to proof this
    ##(n+1)nPr=(n+1)P(r+1)##

    The attempt at a solution
    if i substitute this in
    ##nPr=\frac{n!}{(n-r)!}##
    I get this
    ##(n+1)nPr=\frac{[(n+1)n]!}{[(n+1)n-r]!}##
    This will give
    ##(n+1)nPr=\frac{(n^2+n)!}{(n^2+n-r)!}##

    Now i don't know how to continue.
     
  2. jcsd
  3. Oct 24, 2015 #2

    SteamKing

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    I don't think (n + 1) * n! = [n * (n + 1)]!

    What is (n + 1)*n! ?
     
  4. Oct 24, 2015 #3

    haruspex

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    Further to SteamKing's point, why did you also attempt to multiply the denominator by n+1?
     
  5. Oct 25, 2015 #4
    I just substituted (n+1)n in the equation of nPr.
    ##nPr=\frac{n!}{n-r}##
    Just changed the (n) with (n+1)n
    I didn't multipy the denominator.

    But now i think i get you
    Did you mean that (n+1)nPr isn't the same as ((n+1)n)Pr ??
     
  6. Oct 25, 2015 #5

    haruspex

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    Yes. It's (n+1)(nPr).
     
  7. Oct 25, 2015 #6
    Ok thanks. Now lets go on
    ##(n+1)nPr=(n+1)\frac{n!}{(n-r)!}##

    I have to get this
    ##\frac{(n+1)!}{[(n+1)-(r+1)]!}##
    to proof the R.H.S

    But if i start with the Right side
    ##(n+1)P(r+1)=\frac{(n+1)!}{[(n+1)-(r+1)]!}=(n+1)\frac{n!}{(n-r)!}=(n+1)nPr=L.H.S##
    But how to start with left side and proof the right side ?

    I can -1 +1 to the denominator to get the (n+1)-(r+1)
    But can i say that (n+1)n!=(n+1)! ?
    I think yes of course
     
  8. Oct 25, 2015 #7

    haruspex

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    Yes. And since you are dealing with equalities, not inequalities, it doesn't matter which side you start with to arrive at the other.
     
  9. Oct 25, 2015 #8
    Ok. Thank you very much for your help.
     
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