Compute the number of ordered quadruples of positive integers $(a,b,c,d)$ such that $$a!\cdot b!\cdot c!\cdot d!=24!$$
[list=1]
[*] 4
[*] 4!
[*] $4^4$
[*] None of these
[/list]
To solve the problem, we need to find the number of ordered quadruples \((a, b, c, d)\) of positive integers such that \(a! \cdot b! \cdot c! \cdot d! = 24!\).

1. **Prime Factorization of \(24!\)**:
   \[
   24! = 2^{22} \cdot 3^{10} \cdot 5^4 \cdot 7^3 \cdot 11^2 \cdot 13 \cdot 17 \cdot 19 \cdot 23
   \]

2. **Constraints on \(d\)**:
   - Since \(d\) is the largest number in the quadruple, it must be at least 23. This is because if \(d < 23\), then \(\nu_{23}(a!b!c!d!) = 0\), which contradicts the prime factorization of \(24!\).
   - If \(d = 23\), then:
     \[
     a! \cdot b! \cdot c! = \frac{24!}{23!} = 24
     \]
   - If \(d = 24\), then:
     \[
     a! \cdot b! \cdot c! = \frac{24!}{24!} = 1
     \]

3. **Case \(d = 23\)**:
   - We need \(a! \cdot b! \cdot c! = 24\).
   - The possible factorizations of 24 into factorials are:
     - \(24 = 4!\)
     - \(24 = 3! \cdot 2! \cdot 2!\)
   - Therefore, the solutions are:
     - \((a, b, c) = (1, 1, 4)\)
     - \((a, b, c) = (2, 2, 3)\)

4. **Case \(d = 24\)**:
   - We need \(a! \cdot b! \cdot c! = 1\).
   - The only way to achieve this is:
     - \((a, b, c) = (1, 1, 1)\)

5. **Counting the Ordered Quadruples**:
   - For \(d = 23\):
     - \((a, b, c) = (1, 1, 4)\) and \((a, b, c) = (2, 2, 3)\)
     - Each of these can be permuted in \(\frac{3!}{2!} = 3\) ways (since two elements are the same).
   - For \(d = 24\):
     - \((a, b, c) = (1, 1, 1)\)
     - This can only be permuted in 1 way.

6. **Total Number of Ordered Quadruples**:
   - For \((1, 1, 4, 23)\): \(3\) permutations.
   - For \((2, 2, 3, 23)\): \(3\) permutations.
   - For \((1, 1, 1, 24)\): \(1\) permutation.
   - Total: \(3 + 3 + 1 = 7\)

The final answer is \(\boxed{None \text{ of these}}\).