Question

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

Answer 1

We have,

$\frac{{}^{56}{P}_{r+6}}{{}^{54}{P}_{r+3}}=\frac{30800}{1}$

$\frac{\frac{56!}{(56-r-6)!}}{\frac{54!}{(54-r-3)!}}=\frac{30800}{1}$

$\frac{\frac{56\times 55\times 54!}{(50-r)!}}{\frac{54!}{(51-r)!}}=\frac{30800}{1}$

$\frac{3080\times (51-r)!}{(50-r)!}=30800$

$\frac{(51-r)(50-r)!}{(50-r)!}=10$

$51-r=10$

$r=41$

So, the value of $r$ is $41$.