If P r n = P r + 1 n and C r n = C r - 1 n , then the value of r is equal to:

We have, ${}^{n}P_r={}^{n}P_{r+1}$

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

$\Rightarrow \left(n-r\right)=1 \dots \left(1\right)$

and ${}^{n}C_r={}^{n}C_{r-1}$

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

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

$\Rightarrow n-r+1=r$

$\Rightarrow n+1=2r \dots \left(2\right)$

From equation $\left(1\right) \& \left(2\right),$ we get

$2r-1-r=1$

$\Rightarrow r=2.$