Problem 1212

To solve this problem, we need to determine the greatest integer $k$ such that in any selection of 6 subsets of 5 elements each from the set $\{1, 2, \dots, 9\}$, there exist $k$ subsets that have at least one common element.

1. Calculate the total number of subsets:
The set $\{1, 2, \dots, 9\}$ has $\binom{9}{5}$ subsets of 5 elements.
$$\binom{9}{5} = \frac{9!}{5!(9-5)!} = \frac{9!}{5! \cdot 4!} = \frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1} = 126$$
So, there are 126 possible subsets of 5 elements each.

2. Apply the Pigeonhole Principle:
We need to select 6 subsets out of these 126. By the Pigeonhole Principle, if we distribute 30 elements (since each subset has 5 elements and there are 6 subsets, $6 \times 5 = 30$) among 9 possible elements, at least one element must appear in at least:
$$\left\lceil \frac{30}{9} \right\rceil = 4 \text{ subsets}$$
This means that there is at least one element that appears in at least 4 of the 6 subsets.

3. Construct an example to verify:
We need to check if it is possible to have 6 subsets where no element appears in more than 3 subsets. If we can construct such an example, then $k = 4$ is the greatest integer satisfying the condition.

Consider the following subsets:
$$\{1, 2, 3, 4, 5\}, \{1, 6, 7, 8, 9\}, \{2, 3, 4, 5, 6\}, \{1, 2, 7, 8, 9\}, \{3, 4, 5, 6, 7\}, \{1, 2, 3, 8, 9\}$$
- Element 1 appears in 3 subsets: $\{1, 2, 3, 4, 5\}, \{1, 6, 7, 8, 9\}, \{1, 2, 7, 8, 9\}$
- Element 2 appears in 4 subsets: $\{1, 2, 3, 4, 5\}, \{2, 3, 4, 5, 6\}, \{1, 2, 7, 8, 9\}, \{1, 2, 3, 8, 9\}$
- Element 3 appears in 4 subsets: $\{1, 2, 3, 4, 5\}, \{2, 3, 4, 5, 6\}, \{3, 4, 5, 6, 7\}, \{1, 2, 3, 8, 9\}$
- Element 4 appears in 3 subsets: $\{1, 2, 3, 4, 5\}, \{2, 3, 4, 5, 6\}, \{3, 4, 5, 6, 7\}$
- Element 5 appears in 3 subsets: $\{1, 2, 3, 4, 5\}, \{2, 3, 4, 5, 6\}, \{3, 4, 5, 6, 7\}$
- Element 6 appears in 2 subsets: $\{2, 3, 4, 5, 6\}, \{3, 4, 5, 6, 7\}$
- Element 7 appears in 2 subsets: $\{1, 6, 7, 8, 9\}, \{3, 4, 5, 6, 7\}$
- Element 8 appears in 2 subsets: $\{1, 6, 7, 8, 9\}, \{1, 2, 3, 8, 9\}$
- Element 9 appears in 2 subsets: $\{1, 6, 7, 8, 9\}, \{1, 2, 3, 8, 9\}$

From this construction, we see that it is possible to have 6 subsets where no element appears in more than 4 subsets. Therefore, the greatest integer $k$ is 4.

$\blacksquare$

The final answer is $\boxed{4}$