Question

Prove that:
${}^n{P}_r{=}^{n-1}{P}_r+r{.}^{n-1}{P}_{r-1}$

Answer 1

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

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

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

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

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

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

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

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

${=}^n{P}_r$

Hence proved.