If $^{n} P_{7} = 42. (^{n} P_{5})$ , then find $n$ .

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Solution

$^{n} P_{7} = 42. (^{n} P_{5})$
$\frac{n!}{(n - 7)!} = 42. \frac{n!}{(n - 5)!}$
$\frac{(n - 5)!}{(n - 7)!} = 42$
$(n - 5) (n - 6) = 42 ⟹ n^{2} - 11 n + 30 = 42 ⟹ n^{2} - 11 n - 12 = 0 ⟹ n = - 1, 12$
But $n$ can never be negative So $n = 12$

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