Question

n!/(n-2)! + 25 = (2n)!/ 2(2n-2)!

Answer 1

n!/(n-2)!+25=(2n)!/2(2n-2)!
n(n-1)(n-2)!/(n-2)!+25=(2n)(2n-1)(2n-2)!/2(2n-2)!
n²-n+25=4n²-2n/2
n²-n+25=2n²-n
-n²+25=0
n²=25
n=+-5

mas não existe fatorial negativo

portanto
n=5

$\boxed{ \mathsf{n = 5}}$