Question

Prove that npr=n-1pr+r. (n-1)pr-1.​

Answer 1

Step-by-step explanation:

RHS

(n-1)!/(n-1-r)!+r.(n-1)!/(n-r)!

=(n-1)!/(n-r-1)!+r.(n-1)!/(n-r).(n-r-1)!

=(n-1)!/(n-r-1)!+(1+r/n-r)

=(n-1)!/(n-r-1)!(n/n-r)

=n!/(n-r)!

=nPr

Answer 2

$nP_r=^{n-1}P_r+r\cdot^{n-1}P_{r-1}$ proved.

Step-by-step explanation:

Consider the provided information.

We need to prove that $^nP_r=^{n-1}P_r+r\cdot^{n-1}P_{r-1}$

Consider the Right hand side.

$\begin{aligned}^{n-1}P_r+r\cdot^{n-1}P_{r-1}&=\frac{(n-1)!}{(n-r-1)!}+\frac{r(n-1)!}{(n-r+1-1)!}\\&=\frac{(n-1)!}{(n-r-1)!}+\frac{r(n-1)!}{(n-r)!}\\&=\frac{(n-1)!}{(n-r-1)!}\left[1+\frac{r}{n-r}\right]\\&=\frac{(n-1)!}{(n-r-1)!}\left[\frac{n-r+r}{n-r}\right]\\&=\frac{n(n-1)!}{(n-1)\cdot(n-r-1)!}\\&=\frac{n!}{(n-r)!}\\&=^nP_r\end{aligned}$

Hence, proved

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