Question

If ${}^n{P}_r=5040({}^{n-1}{C}_{r-1}+{}^{n-1}{C}_r)$,then r =

• A. 6
• B. 7
• C. 5
• D. None

Answer 1

The correct option is B

7

${}^n{P}_r=5040({}^{n-1}{C}_{r-1}+{}^{n-1}{C}_r)=5040{}^n{C}_r$

$\Rightarrow$ $\frac{n!}{(n-r)!}=5040\frac{n!}{(n-r)r!}\Rightarrow r!=5040$

$\Rightarrow$ r = 7.