Question

If ${}^n{P}_r={}^n{P}_{r+1}$ and ${}^n{C}_r={}^n{C}_{r-1}$, then the value of $r$ is equal to

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

Given: ${}^n{P}_r={}^n{P}_{r+1}$
$\Rightarrow \frac{n!}{(n-r)!}=\frac{n!}{(n-r-1)!}$
$\Rightarrow (n-r)(n-r-1)!=(n-r-1)!$
$\therefore n-r=1\cdots \left(1\right)$
And ${}^n{C}_r={}^n{C}_{r-1}$
$\Rightarrow \frac{n!}{r!\cdot (n-r)!}=\frac{n!}{(n-r+1)!\cdot (r-1)!}$
$\Rightarrow r(r-1)!\cdot (n-r)!=(n-r+1)(n-r)!\cdot (r-1)!$
$\Rightarrow r=n-r+1$
$\Rightarrow r=1+1$ $\left(\text{From}\right(1\left)\right)$
$\therefore r=2,n=3$