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

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### 3 Answers

We are given that n! / (n - 2)! = 342 and we have to find n.

Now n! is equal to the product of all integers from 1 to n. Or n!=1*2*3...n.

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

Therefore (n-2)! does not have the terms n and n-1 that n! does.

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

=> n^2 + n = 342

=> n^2 + n - 342 =0

=> n^2 + 19n - 18n - 342 =0

=> n(n+19) - 18(n+19) =0

=> (n-18) (n+19) =0

So n can be equal to -18 or 19. But the factorial of a negative number is not defined. Also, it is given that n is positive, therefore we take n as 19.

**The required value of n is 19.**

Give n!/(n-2)! = 342, where n is a positive integer.

We know for postinve integers we can write n! = n*(n-1)(n-2!.

Therefore n!/(n-2)! = n(n-1)*(n-2)!/(n-2)! = n(n-1).

Therefore n(n-1) = 342

So n^2-n -342 = 0.

n^2- 19n +18n -342 = 0.

n(n-19) +18(n -19) = 0.

(n-19)(n+18) = 0.

Therefore n -19 = 0, or n =19 is the solution.

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.**