Question

PROVE THAT nPr=(n-r)Pr+r.(n-1)P(r-1)

Answer 1

I hope this answer will help you.

Answer 2

Answer and Explanation:

To prove : $^nP_r=^{n-r}P_r+r\cdot^{n-1}P_{r-1}$

Proof :

We know, $^nP_r=\frac{n!}{(n-r)!}$

Taking LHS,

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

We can write, $n!=n(n-1)!$

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

Add and subtract r,

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

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

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

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

$^nP_r=\frac{(n-1)!}{(n-1-r)!}+r[\frac{(n-1)!}{(n-r)(n-r-1)!}]$

Again form a formula,

$^nP_r=^{n-1}P_r+r[\frac{(n-1)!}{(n-r)!}]$

$^nP_r=^{n-1}P_r+r[\frac{(n-1)!}{(n-1-(r-1)!}]$

$^nP_r=^{n-1}P_r+r[^{n-1}P_{r-1}}]$

Hence proved.