# n = ?If n is a positive integer so n! / (n - 2)! = 342, find n.

### 2 Answers | Add Yours

You need to cross multiply, such that:

`n! = 342(n - 2)!`

You need to use the equation that relates factorials, such that:

`n! = (n-1)!*n = (n - 2)!*(n - 1)*n = (n - 3)!*(n - 2)(n - 1)n = ... = 1*2*3*..*(n - 2)(n - 1)*n`

Hence, substituting `(n - 2)!*(n - 1)*n` for `n!` , yields:

`(n - 2)!*(n - 1)*n = 342(n - 2)!`

Reducing duplicate factors, yields:

`n(n - 1 ) = 342 => n^2 - n = 342`

You need to complete the square `n^2 - n` , such that:

`n^2 - n + 1/4 = 342 + 1/4`

`(n - 1/2)^2 = 1369/4 => n - 1/2 = +-sqrt(1369/4) => n - 1/2 = +-37/2`

`n = 1/2 +- 37/2 => n = 19 ; n = -18`

**The result `n = -18` is invalid because `-18!in N` , hence, evaluating the solution to the given factorial equation yields **`n = 19.`

We know that n! = (n - 2)!*(n-1)*n (1)

We'll substitute (1) into the given equation:

(n - 2)!*(n-1)*n/ (n - 2)! = 342

We'll simplify and we'll get:

n(n-1) = 342

We'll remove the brackets and we'll get:

n^2 - n = 342

We'll subtract 342 both sides:

n^2 - n - 342 = 0

We'll apply the quadratic formula:

n1 = [1 + sqrt(1 +1368 )]/2

n1 = (1+37)/2

n1 = 19

n2 = (1-37)/2

n2 = -18

Since n is a positive integer, we'll reject the second solution and we'll keep just n = 19.