If $(n + 2)! = 60 (n - 1)!$ find n.

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Solution

We have,
$(n + 2)! = 60 (n - 1)$

As we know $n! = (n - 1) n!$

Therefore,
$(n + 2) (n + 1) n \times (n - 1)! = 60 (n - 1)!$
$(n + 2) (n + 1) n = 60$
$(n^{2} + 2 n + n + 2) n = 60$
$(n^{3} + 3 n^{2} + 2 n - 60) = 0$

Factoring the left hand side, this equation resolves to
$(n - 3) (n^{2} + 6 n + 20) = 0$

Since it can be shown that has $(n^{2} + 6 n + 20) = 0$ no real solution,
$n - 3 = 0$
$n = 3$

Hence, the value of $n$ is $3$ .

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