Question

If $\rm ^{11}P_{r-1} : \rm ^{12}P_{r-2} = 14 : 3$, find r.

Answer 1

Solution:

As we know that

$\rm ^{n}P_{r} = \frac{n!}{(n - r)!}$
$\rm^{11}P_{r-1} : \rm ^{12}P_{r-2} = 14 : 3 \\\\ \frac{\rm ^{11}P_{r-1}}{\rm ^{12}P_{r-2}} = \frac{14}{3} \\ \\ \frac{11!}{(11 - r + 1)!} \times \frac{(12 - r + 2)!}{12!} = \frac{14}{3} \\ \\ \frac{11!(14 - r)!}{12!(12 - r)!} = \frac{14}{3} \\ \\ \frac{(14 - r)(13 - r)}{12} = \frac{14}{3} \\ \\ (14 - r)(13 - r) = 14 \times 4 \\ \\ 182 - 14r - 13r + {r}^{2} = 56 \\ \\ {r}^{2} - 27r + 126 = 0 \\ \\ solve \: this \: for \: quadratic \: formula \\ \\ r_{1,2} = \frac{27 ± \sqrt{729 - 504} }{2} \\ \\ r_{1,2} = \frac{27 ± \sqrt{225} }{2} \\ \\r_{1,2}=\frac{27 ± 15}{2} \\ \\ r_{1} = \frac{27 + 15}{2} \\ \\ r_{1} = 21 \\ \\ r_{2} = \frac{27 - 15}{2} \\ \\ r_{2} = 6$

Thus r can be 21 and 6.

Hope it helps you.