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 -...

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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**