We have to solve for n given that : (2n)!=30(2n-2)!

The factorial function n! = 1*2*...*n

(2n)!=30(2n-2)!

=> 2n*(2n - 1)(2n - 2)! = 30*(2n - 2)!

=> 2n*(2n - 1) = 30

=> n(2n - 1) = 15

=> 2n^2 - n - 15 = 0

=> 2n^2 - 6n + 5n - 15 = 0

=> 2n(n - 3) + 5(n - 3) = 0

=> (2n + 5)(n - 3) = 0

but factorial is defined only for positive integers.

**So we have n = 3**

The factorial of a number n, n! is equal to the product n*(n-1)*(n-) ...1

The equation (2n)!=30(2n-2)! has to be solved for n.

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

Substitute this in the equation

(2n)*(2n-1)*(2n-2)! = 30*(2n-2)!

Cancel (2n-2)!

2n*(2n-1) = 30

Cancel 2

n*(2n-1) = 15

2n^2 - n - 15 = 0

2n^2 - 6n + 5n - 15 = 0

2n(n - 3) + 5(n - 3) = 0

(2n + 5)(n - 3) = 0

This gives n = -5/3 and n = 3

The factorial is defined only for positive numbers, as a result the root n = -5/3 can be eliminated.

This gives the required value of n as 3.

By definition, (2n)! = (2n-2)!*(2n-1)*2n

We'll re-write the equation:

(2n-2)!*(2n-1)*2n = 30(2n-2)!

We'll divide by (2n-2)!:

(2n-1)*2n = 30

We'll divide by 2:

n(2n - 1) = 15

We'll remove the brackets:

2n^2 - n - 15 = 0

We'll apply quadratic formula:

n1 = [1 + sqrt(1 + 120)]/4

n1 = (1 + 11)/4

n1 = 3

n2 = -10/4

n2 = -5/2

**Since just n1= 3 is an element of N set, we'll reject the value of n2. The equation has only one solution, n = 3.**