If $^{n} P_{5} = 20^{n} P_{3},$ , find the value of $n$ .

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Solution

$G i v e n,$
$P_{5}^{n}$ $=$ $20$ $P_{3}^{n}$
$\frac{n!}{(n - 5)!}$ $=$ $20$ $\frac{n!}{(n - 3)!}$
$\frac{n!}{(n - 5)!}$ $=$ $20$ $\frac{n!}{(n - 3) (n - 4) (n - 5)!}$
$1$ $=$ 20 $\frac{1}{(n - 3) (n - 4)}$
$(n - 3) (n - 4) = 20$
$n^{2} - 7 n + 12 = 20$
$n^{2} - 7 n + 12 - 20 = 0$
$n^{2} - 7 n - 8 = 0$
$n^{2} + n - 8 n - 8 = 0$
$n (n + 1) - 8 (n + 1) = 0$
$(n - 8) (n + 1) = 0$
$n = 8, - 1$
Since , n can not be nagative . $n \neq$ $- 1$
Hence, $n = 8.$

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