Question

If ${}^{18}{P}_{r-1}{:}^{17}{P}_{r-1}=9:7$ then find $r$.

Answer 1

As we know that,

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

Therefore,

${}^{18}{P}_{r-1}=\frac{18!}{(18-(r-1))!}=\frac{18!}{(19-r)!}$

${}^{17}{P}_{r-1}=\frac{17!}{(17-(r-1))!}=\frac{17!}{(18-r)!}$

Now,

$\frac{{}^{18}{P}_{r-1}}{{}^{17}{P}_{r-1}}=\frac{9}{7}$

$\frac{\left(\frac{18!}{(19-r)!}\right)}{\left(\frac{17!}{(18-r)!}\right)}=\frac{9}{7}$

$\frac{\left(\frac{18\times 17!}{(19-r)(18-r)!}\right)}{\left(\frac{17!}{(18-r)!}\right)}=\frac{9}{7}$

$\Rightarrow \frac{18}{19-r}=\frac{9}{7}$

$\Rightarrow 19-r=14$

$\Rightarrow r=19-14=5$