Do Inclusion-Exclusion and Pigeonhole Principles Apply to Overlapping Sets?

[fiksx]

Homework Statement

If the proposition below is true, show this. If false, give a counterexample.
From the integers from 1 to 11, make 10 sets $S_1,S_2, \dots, S_{10}$ each with 4 integers selected. Within each sets, the same number shall not be chosen twice. No matter how $S_1,S_2, \dots, S_{10}$ is selected, two sets $S_i,S_j$ (i not equal to j) include two or more common integers.

Relevant Equations

Sets

Is this related to pigeon principle?
$$S_1=\{1,2,3,4\},$$
$$S_2=\{2,3,4,5\},$$
$$S_3=\{4,5,6,7\},$$
$$S_4=\{5,6,7,8\},$$
$$S_5=\{7,8,9,10\},$$
$$S_6=\{8,9,10,11\},$$
$$S_7=\{5,6,2,4\},$$
$$S_8=\{1,5,7,9\},$$
$$S_9=\{4,8,10,11\},$$
$$S_{10}=\{5,7,10,11\}$$

When we choose two of them, there is possibility there are same integer but not all ?
If this is related to pigeonhole principle or inclusion exclusion principle, Is there possibility that sets are hole and number 1-11 is pigeon but what is the relation with in each sets there are four number?

 

[Orodruin]

The language in your problem statement is ambiguous. It is not clear whether the proposition is:

$\exists i \neq j$: $S_i$ and $S_j$ have two or more common elements.

or

$\forall i \neq j$: $S_i$ and $S_j$ have two or more common elements.

Your sets clearly is a counter example to the second, but not to the first.

 

[fiksx]

I also thought about that, but if it is like second, it is too obvious, but if it is like the first , how should i proof that? Is it use inclusion exclusion principle?