Find the value of n such that
$^{n} P_{5} = 42$ $^{n} P_{3}$ , $n > 4$

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Solution

Given that: $^{n} P_{5} = 42 \times^{n} P_{3}$
$=> n (n - 1) (n - 2) (n - 3) (n - 4) = 42 n (n - 1) (n - 2) ∵ n > 4, s o n (n - 1) (n - 2) \neq 0$
Therefore, by dividing both the sides with $n (n - 1) (n - 2)$ , we get,
$(n - 3) (n - 4) = 42 => n^{2} - 7 n - 30 = 0 => n^{2} - 10 n + 3 n - 30 = 0 => n (n - 10) + 3 (n - 10) = 0 => (n + 3) (n - 10) = 0 => n = - 3, 10$
As $n$ cannot be negative, so $n = 10$ .

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