Two permutations $a_1,a_2,\cdots,a_{2010}$ and $b_1,b_2,\cdots,b_{2010}$ of the numbers $1,2,\cdots,2010$ are said to intersect if $a_k=b_k$ for some value $k$ in the range $1\le k \le 2010$. Show that there exist $1006$ permutations of the numbers $1,2,\cdots,2010$ such that any other such permutation is guaranteed to intersect at least one of these $1006$ permutations.
Well, I have the feeling that I should approach this using the Pigeonhole Principle, but I don't have any idea how to start.
Any help would surely be appreciated. Thanks!
HINT
There exist $3$ permutations, call them $\sigma_1, \sigma_2, \sigma_3$, of the numbers $\{1,2,3,4\}$ s.t. any other permutation $\pi$ intersects at least one of these 3 $\sigma_i$ permutations. Can you construct such 3 permutations?
Further hint: Consider the value $1$. It has to be in some position in $\sigma_1$, and whatever position it's in, that would prevent $\pi$ from having $1$ in the same position (or else they would intersect). Now you have $3$ different $\sigma$'s and therefore can "block out" up to $3$ different positions for the value $1$, so $\pi$ must have the value $1$ in the remaining position. But you haven't used value $2$ for blocking...
Can you finish from here?