Question

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

Answer 1

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

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

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

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

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

$\Rightarrow$ $\frac{(n-1)!}{r!\left(\right(n-1)-r)!}={}^{n-1}{P}_r$

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

Hence, the answer is ${}^{n-1}{P}_r.$