Question

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

• A. $1640$
• B. $1890$
• C. $1840$
• D. $1680$

Answer 1

Answer: (A)

$1640$

Consider LHS

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

But LHS $=$ RHS

$\therefore 56\times 55\times (51-r)=30800\therefore 3080\times (51-r)=30800\therefore 51-r=10\therefore r=41$

${\therefore }^r{P}_2{=}^{41}{P}_2=\frac{41!}{(41-2)!}=\frac{41\times 40\times 39!}{39!}=41\times 40=1640$

Hence, the answer is option A.