Task B-4.1. Determine the natural number $n \geq 2$ such that the equality

$$
\frac{(n-1)^{2} n(n+1)!}{(n+2)!}=\binom{n}{2}
$$

holds.
$$
\begin{aligned}
\frac{(n-1)(n-1) n(n+1)!}{(n+2)!} & =\frac{n!}{2!(n-2)!} \\
\frac{(n-1)(n-1) n(n+1)!}{(n+2)(n+1)!} & =\frac{(n-2)!(n-1) n}{2(n-2)!} \\
\frac{n-1}{n+2} & =\frac{1}{2} \\
2 n-2 & =n+2 \\
n & =4
\end{aligned}
$$