Question

If (n + 2)! = 60 (n − 1)!. Find n.

Answer 1

Hint:
The factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n.
 As we know n! = (n − 1)n!.
Therefore, $\left( {n + 2} \right)\left( {n + 1} \right)n\left( {n - 1} \right)! = 60\left( {n - 1} \right)!$
⇒ ${n^3} + 3{n^2} + 2n - 60 = 0$
Then proceed to solve it by factorization method.

Complete step by step solution:
The factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n .
Give expression,
(n + 2)! = 60 (n − 1)!
⇒ As we know that,
$n! = (n − 1)n!$
Therefore,
 $\left( {n + 2} \right)\left( {n + 1} \right)n\left( {n - 1} \right)! = 60\left( {n - 1} \right)!$
⇒ After dividing and multiplying we get,
 $ \Rightarrow \left( {n + 2} \right)\left( {n + 1} \right)n = 60$
⇒ $\left( {{n^2} + 2n + n + 2} \right)n = 60$
⇒ ${n^3} + 3{n^2} + 2n = 60$
⇒ ${n^3} + 3{n^2} + 2n - 60 = 0$
Factoring the left-hand side, this equation resolves to,
⇒ $\left( {n - 3} \right)\left( {{n^2} + 6n + 20} \right) = 0$

either.$\left( {n - 3} \right) = 0\;or\;{n^2} + 6n + 20 = 0$
but ${n^2} + 6n + 20 = 0\;$ has no real solution because their discriminant is less than zero.
⇒ ${b^2} - 4ac < 0$
Now,
$n − 3 = 0$
$⇒ n = 3$

Note:
In this type of question first we simplify the equation by writing factorial as the product. Then we use factorization to find the value of n.