1. ## 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

2. In general $\displaystyle P(n,j) = \frac{{n!}}{{\left( {n - j} \right)!}}$ so we have:
$\displaystyle \begin{gathered} P(n,2) = \frac{{n!}} {{\left( {n - 2} \right)!}} \hfill \\ P(n - 1,2) = \frac{{\left( {n - 1} \right)!}} {{\left( {n - 3} \right)!}} \hfill \\ P(n - 1,1) = \frac{{\left( {n - 1} \right)!}} {{\left( {n - 2} \right)!}} \hfill \\ \end{gathered}$.

3. thanks A LOT! for getting me started i have figured it out!

they both reduce to (n^2) - n

4. Originally Posted by DaCoo911
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$\displaystyle P(n,r)=\frac{n!}{(n-r)!}$

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

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

$\displaystyle \frac{(n-1)(n-2)![n-2]}{(n-2)(n-3)!}=\frac{(n-1)!}{(n-3)!}$

$\displaystyle (n-1)(n-2)!=(n-1)!$