Question

If ${}^{56}{P}_{r+6}:{}^{54}{P}_{r+3}=30800:1$, find ${}^r{P}_2$.

Answer 1

We know that,

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

Now,

${}^{56}{P}_{r+6}{:}^{54}{P}_{r+3}={10.}^{56}{P}_2:1\Rightarrow \frac{56!}{(50-r)!}÷\frac{54!}{(51-r)!}=10.\frac{56!}{54!}\Rightarrow \frac{56!}{54!}\times \frac{(51-r)!}{(50-r)!}=10\ast 56\times 55\Rightarrow 56\ast 55(51-r)=10\ast 56\ast 55\Rightarrow 51-r=10\Rightarrow r=41$

Now, ${}^{41}{P}_2=\frac{41!}{39!}=41\ast 40=1640$