Compute the number of ordered quadruples of positive integers $(a,b,c,d)$ such that
$$a!b!c!d!=24!$$$\textbf{(A)}~4$
$\textbf{(B)}~4!$
$\textbf{(C)}~4^4$
$\textbf{(D)}~\text{None of the above}$
To solve the problem, we need to find the number of ordered quadruples \((a, b, c, d)\) of positive integers such that \(a!b!c!d! = 24!\).

1. **Factorial Decomposition**:
   First, we note that \(24! = 24 \times 23 \times 22 \times \cdots \times 1\). Since \(24!\) is a very large number, we need to consider the possible values for \(a, b, c,\) and \(d\) such that their factorials multiply to \(24!\).

2. **Largest Factorial**:
   One of the numbers must be large enough to account for most of the factorial product. The largest factorial less than or equal to \(24!\) is \(23!\). Therefore, one of \(a, b, c,\) or \(d\) must be \(24\) (since \(24! = 24 \times 23!\)).

3. **Remaining Factorials**:
   After choosing one of the numbers to be \(24\), we need to find the remaining three numbers such that their factorials multiply to \(1\) (since \(24! / 24! = 1\)). The only factorial that equals \(1\) is \(0!\) or \(1!\). Since we are dealing with positive integers, the only possibility is \(1!\).

4. **Possible Combinations**:
   Therefore, the remaining three numbers must all be \(1\). This gives us the quadruple \((24, 1, 1, 1)\).

5. **Permutations**:
   Since the problem asks for ordered quadruples, we need to count the permutations of \((24, 1, 1, 1)\). The number of permutations of this set is given by the multinomial coefficient:
   \[
   \frac{4!}{1! \cdot 3!} = \frac{24}{6} = 4
   \]

Thus, there are 4 ordered quadruples \((a, b, c, d)\) that satisfy the equation \(a!b!c!d! = 24!\).

The final answer is \(\boxed{4}\).