Question

If ${}^{56}{P}_{R+6}{:}^{54}{P}_{R+3}=30800:1$ Find ${}^R{P}_2.$

Answer 1

$\frac{\frac{56!}{(56-R-6)!}}{\frac{54!}{(54-R-3)!}}$ = $\frac{\frac{56\times 55\times 54!}{(50-R)!}}{(51-R)(50-R)!}$

$\Rightarrow 56\times 55\times (51-R)=30,800$

51-R=10 $\Rightarrow R=41$

$\therefore {41}_{P_2}=\frac{41!}{(41-2)!}=41\times 40=1640$