Question

If ${^{n + 2}C_8} : {^{n - 2}P_4} = 57 : 16,$ then the value of n is:

• A. 20
• B. 19
• C. 18
• D. 17

Answer 1

Answer: (B)

Given $^{n+2} C_{8} : ^{n-2}P_{4} = 57 : 16$
$\frac{^{n+2}C_{8}}{^{n-2}P_{4}} = \frac{57}{16} \left[\because ^{n}C_{r} = \frac{n!}{r! \left(n-r\right)!} \text{and} ^{n}P_{r} = \frac{n!}{\left(n-r\right)!}\right]$
$\Rightarrow \frac{\left(n+2\right)!}{8!\left(n+2-8\right)!} \times \frac{\left(n-2-4\right)!}{\left(n-2\right)!} = \frac{57}{16}$
$\Rightarrow \frac{\left(n+2\right)\left(n+1\right)n.\left(n-1\right)}{8.7.6.5.4.3.2.1} = \frac{57}{16}$
$\Rightarrow \left(n + 2\right) \left(n + 1\right) n \left(n - 1\right) = 143640$
$\Rightarrow \left(n^{2} + n - 2\right) \left(n2 + n\right) = 143640$
$\Rightarrow \left(n^{2} + n\right)^{2} - 2\left(n^{2} + n\right) + 1 = 143640 + 1$
$\Rightarrow \left(n^{2} + n - 1\right)^{2} = \left(379\right)^{2}$
$\Rightarrow n^{2} + n - 1 = 379 \left[\because n^{2} + n - 1 > 0\right]$
$\Rightarrow n^{2} + n - 1 - 379 = 0$
$\Rightarrow n^{2} + n - 380 = 0$
$\Rightarrow \left(n + 20\right) \left(n - 19\right) = 0$
$\Rightarrow n = - 20, n = 19$
$\because$ n is not negative.
$\therefore n = 19$