Hint/Spoiler for Problem 4 on Assignment 1

This is a non-obvious application of the Pigeonhole Principle.

Objects: Integers 1,2,...,10

Boxes: All ordered pairs of positive integers: {(i,j):i,j >= 1}

Place integer a (which appears in position k, say, in the permutation) in box (i,j) where i is the length of the longest increasing subsequence ending at a and where j is the length of the longest decreasing subsequence ending at a.

For example, if the permutation is 5 1 4 8 2 6 3 10 9 7, then 5 goes in box (1,1), 1 goes in box (1,2), 4 goes in box (2,2), and so on.

No two integers can go into the same box. (Why?)

So some box with i>3 or j>3 must get filled.