Let \( S_n = 1 \cdot (n - 1) + 2 \cdot (n - 2) + 3 \cdot (n - 3) + \ldots + (n - 1) \cdot 1, \; n \ge 4. \) The sum \( \displaystyle \sum_{n=4}^{\infty}\left( \frac{2 S_n}{n!} - \frac{1}{(n-2)!} \right) \) is equal to:
\( \frac{e}{6} \)
\( \frac{e-1}{3} \)
\( \frac{e}{3} \)
\( \frac{e-2}{6} \)