Posting the part of the solution...i know the concept on which it is asked but i got stuck while solving and i peeked into the solution, i found this
$=\frac{2n(2n-1)(2n-2)...4.3.2.1}{n!n!}x^n$ fine $=\frac{1.2.3.4...(2n-2)(2n-1)(2n)}{n!n!}x^n$ fine $=\frac{[1.3.5...(2n-1)][2.4.6...(2n)]}{n!n!}x^n$ fine $=\frac{[1.3.5..(2n-1)]2^n[1.2.3...n]}{n!n!}x^n$ 2^n is taken as common, i'm tricked here, n is raised to 2 which means there are n terms into $“[2.4.6...(2n)]”$ but how? Please explain how there are n terms into that expression ($“[2.4.6...(2n)]”$)
$\begin{align} 2×4×\cdots ×(2n)&=(2×1)×(2×2)×(2×3)×\dots×(2×n)\\ &=2^n×(1×2×\cdots×n ) \end{align}$