2. For a positive integer $n$, let $1 \times 2 \times \cdots \times n=n$!. If $M=1!\times 2!\times \cdots \times 10!$, then the number of perfect cubes among the positive divisors of $M$ is ( ).
(A) 468
(B) 684
(C) 846
(D) 648
2. A.

Notice that,
$$
\begin{array}{l}
M=1! \times 2! \times \cdots \times 10! \\
=2^{9} \times 3^{8} \times 4^{7} \times 5^{6} \times 6^{5} \times 7^{4} \times 8^{3} \times 9^{2} \times 10 \\
=2^{9+2 \times 7+5+3 \times 3+1} \times 3^{8+5+2 \times 2} \times 5^{6+1} \times 7^{4} \\
=2^{38} \times 3^{17} \times 5^{7} \times 7^{4} .
\end{array}
$$

A perfect cube number \( n \) (where \( n \mid M \)) should have the form
$$
n=2^{3 x} \times 3^{3 y} \times 5^{3 z} \times 7^{3 u} (x, y, z, u \in \mathbf{N}),
$$

and \( 3 x \leqslant 38, 3 y \leqslant 17, 3 z \leqslant 7, 3 u \leqslant 4 \).
Therefore, such \( n \) has a total of \( 13 \times 6 \times 3 \times 2 = 468 \) numbers.