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Math Help - Permutations

  1. #1
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    Exclamation Permutations

    I've just started learneing about permutations and i've come across a problem that has completely stumped me.

    Prove P(n, 2) = P(n-1, 2) + 2P(n-1, 1)

    Thanks, Joe
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  2. #2
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    In general P(n,j) = \frac{{n!}}{{\left( {n - j} \right)!}} so we have:
    \begin{gathered}<br />
  P(n,2) = \frac{{n!}}<br />
{{\left( {n - 2} \right)!}} \hfill \\<br />
  P(n - 1,2) = \frac{{\left( {n - 1} \right)!}}<br />
{{\left( {n - 3} \right)!}} \hfill \\<br />
  P(n - 1,1) = \frac{{\left( {n - 1} \right)!}}<br />
{{\left( {n - 2} \right)!}} \hfill \\ <br />
\end{gathered} .
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  3. #3
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    thanks A LOT! for getting me started i have figured it out!

    they both reduce to (n^2) - n
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  4. #4
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    Quote Originally Posted by DaCoo911 View Post
    I've just started learneing about permutations and i've come across a problem that has completely stumped me.

    Prove P(n, 2) = P(n-1, 2) + 2P(n-1, 1)

    Thanks, Joe

    Make use of the fact that  P(n,r)=\frac{n!}{(n-r)!}

    \frac{n!}{(n-2)!}=\frac{(n-1)!}{(n-3)!}+\frac{2(n-1)!}{(n-2)!}

    \frac{n!}{(n-2)!}-\frac{2(n-1)!}{(n-2)!}=\frac{(n-1)!}{(n-3)!}

     <br />
\frac{(n-1)(n-2)![n-2]}{(n-2)(n-3)!}=\frac{(n-1)!}{(n-3)!}<br />

    (n-1)(n-2)!=(n-1)!
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