Answer: D

Solution:

Write $n = 2^a 3^b 5^c 7^d \cdots.$ Since $5! = 2^3 \cdot 3 \cdot 5$ has no other primes, $n$ can only involve $2, 3, 5, 7.$ Matching exponents in $\operatorname{lcm}(5!, n) = 5 \cdot \gcd(10!, n):$

For $2:$ $\max(3, a) = \min(8, a),$ so $3 \le a \le 8$ gives $6$ values. For $3:$ $\max(1, b) = \min(4, b),$ so $1 \le b \le 4$ gives $4$ values.

For $5:$ $\max(1, c) = 1 + \min(2, c)$ with $c \ge 1,$ which forces $c = 3,$ giving $1$ value. For $7:$ $\max(0, d) = \min(1, d),$ so $d = 0$ or $1,$ giving $2$ values.

The total is $6 \cdot 4 \cdot 1 \cdot 2 = 48.$

Thus, D is the correct answer.