Find the value(s) of n: 2P(n, 2)+50 = P(2n, 2)

For 2P(n, 2)+50 I simplified to 2(n)(n-1)+50. For P(2n, 2) I can't get any simpler than $2n\times(2n-1)\times...\times(n+1)\times n\times(n-1)$ and I'm stuck.

BTW what's the LaTex for pick notation?

  • $\begingroup$ If you mean $P(n,2)=\binom n2$, then it is "\binom{n}{2}" between $ signs. $\endgroup$ – Berci Nov 3 '13 at 22:49
  • $\begingroup$ @Berci that's for "combinations", no? $\endgroup$ – Celeritas Nov 3 '13 at 22:55
  • $\begingroup$ @Celeritas yes, you are probably right. $\endgroup$ – Berci Nov 3 '13 at 23:19

If $P(n,2)$ is $n(n-1)$ then you have $$2n(n-1)+50 =2n(2n-1)$$ which is a quadratic you can solve.

You will get a positive and negative answer. Presumably you want the positive integer as your solution.

  • $\begingroup$ I don't follow. Why do you randomly say P(n,2)=n-1? $\endgroup$ – Celeritas Nov 3 '13 at 22:57
  • 1
    $\begingroup$ @Celeritas I did not say $n-1$ but $n(n-1)$. See en.wikipedia.org/wiki/Permutation#In_combinatorics $\endgroup$ – Henry Nov 3 '13 at 22:57
  • $\begingroup$ I don't follow. Why do you randomly say P(n,2)=n(n-1)? $\endgroup$ – Celeritas Nov 3 '13 at 23:05
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    $\begingroup$ I guess then $P(n,k)$ wants to denote the number '$k$-permuations of $n$' which is $n!/(n-k)!$. $\endgroup$ – Berci Nov 3 '13 at 23:22
  • $\begingroup$ @Celeritas - Permutations: choose $k$ items from $n$ where the order in which you choose them matters. So for $P(n,2)$ choose one of the $n$ and then choose one from the remaining $n-1$ in a total of $n \times (n-1)$ different ways. $\endgroup$ – Henry Nov 3 '13 at 23:30

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