Problem: In Duma, there are 1600 delegates, who have formed 16000 committees of 80 persons each. Prove that one can find two committees having at least four common members.
There is a probabilistic solution to this question, which I am having trouble following because I have no background in probability. It begins by calculating the expected number of common members of any two given committees and using this to find the solution. How exactly does this work?
I am also interested to know if anyone has a non-probabilistic solution.
EDIT: After further searching, I found the solution worked out in full detail on page 2 here: http://www.math.cmu.edu/~ploh/docs/math/mop2010/prob-comb-soln.pdf