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If $^{n} P_{7} = {42.}^{n} P_{5},$ then the value of $' n'$ is

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answer is D.

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Detailed Solution

$^{n} P_{7} = {42.}^{n} P_{5}$

$\frac{n!}{(n - 7)!} = 42 \frac{n!}{(n - 5)!}$

$\frac{1}{(n - 7)!} = 42 \times \frac{1}{(n - 5) (n - 6) (n - 7)!}$

$(n - 5) (n - 6) = 7 \times 6$

$(n - 6) (n - 5) = (12 - 6) (12 - 5)$

$n = 12$

$\Rightarrow n - 6 = 6 \Rightarrow n = 12$

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