Question

Prove that:

${}^n{P}_r{=}^{(n-1)}{P}_r+r{}^{(n-1)}{P}_{(r-1)}$

Answer 1

We have,

${}^n{P}_r=\frac{n!}{(n-r)!}$.

Now,

${}^{n-1}{P}_r+r{}^{n-1}{P}_{r-1}$

$=\frac{(n-1)!}{(n-r-1)!}+r.\frac{(n-1)!}{(n-r)!}$

$=\frac{(n-1)!}{(n-r-1)!}\{1+\frac{r}{n-r}\}$

$=\frac{(n-1)!}{(n-r-1)!}\left\{\frac{n}{n-r}\right\}$

$=\frac{\left(n\right)!}{(n-r)!}$

$={}^n{P}_r$