Each row of a $24 \times 8$ table contains some permutation of the numbers $1, 2, \cdots , 8.$ In each column the numbers are multiplied. What is the minimum possible sum of all the products?
[i](C. Wu)[/i]
1. **Understanding the Problem:**
   We have a \(24 \times 8\) table where each row contains a permutation of the numbers \(1, 2, \ldots, 8\). We need to find the minimum possible sum of the products of the numbers in each column.

2. **Using AM-GM Inequality:**
   The Arithmetic Mean-Geometric Mean (AM-GM) Inequality states that for any non-negative real numbers \(a_1, a_2, \ldots, a_n\),
   \[
   \frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \cdots a_n}
   \]
   with equality if and only if \(a_1 = a_2 = \cdots = a_n\).

3. **Applying AM-GM to the Problem:**
   Let \(P_i\) be the product of the numbers in the \(i\)-th column. We need to minimize the sum \(S = P_1 + P_2 + \cdots + P_8\).

   Since each row is a permutation of \(1, 2, \ldots, 8\), the product of all numbers in each row is \(8!\). Therefore, the product of all numbers in the entire table is \((8!)^{24}\).

   By AM-GM, we have:
   \[
   \frac{P_1 + P_2 + \cdots + P_8}{8} \geq \sqrt[8]{P_1 P_2 \cdots P_8}
   \]
   Since \(P_1 P_2 \cdots P_8 = (8!)^{24}\), we get:
   \[
   \sqrt[8]{P_1 P_2 \cdots P_8} = \sqrt[8]{(8!)^{24}} = (8!)^3
   \]
   Therefore,
   \[
   \frac{P_1 + P_2 + \cdots + P_8}{8} \geq (8!)^3
   \]
   which implies
   \[
   P_1 + P_2 + \cdots + P_8 \geq 8 \cdot (8!)^3
   \]

4. **Achieving Equality:**
   To achieve equality in the AM-GM inequality, all \(P_i\) must be equal. We can construct such a table by arranging the rows as follows:
   - First row: \(1, 2, 3, 4, 5, 6, 7, 8\)
   - Second row: \(2, 3, 4, 5, 6, 7, 8, 1\)
   - Third row: \(3, 4, 5, 6, 7, 8, 1, 2\)
   - ...
   - Eighth row: \(8, 1, 2, 3, 4, 5, 6, 7\)
   - Repeat this pattern for the remaining 16 rows.

   In this arrangement, each column contains exactly one of each number from \(1\) to \(8\) three times, so the product in each column is \((8!)^3\).

   Therefore, the sum of the products is:
   \[
   8 \cdot (8!)^3
   \]

The final answer is \(\boxed{8 \cdot (8!)^3}\).