C

$${\sum_{n=1}^{\infty}{\frac{1}{n^2}}=\frac{\pi^2}{6}}$$

$${\frac{\sum_{n=1}^{\infty}\{\prod(_nC_{n-1})\}^2}{(n!)^2}=\frac{\pi^2}{6}}$$

$${\frac{\sum_{n=1}^{\infty}\{\prod(_nC_{n-1})\}^2}{\pi^2}=X}$$

$${\frac{(n!)^2}{6}=Y}$$

$${X=Y}$$

$${\sum_{n=1}^{\infty}{\frac{1}{n^2}}=\frac{\pi^2}{6}}$$   $${?}$$