In a group of students, 50 speak English, 50 speak French and 50 speak Spanish. Some students speak more than one language. Prove it is possible to divide the students into 5 groups (not necessarily equal), so that in each group 10 speak English, 10 speak French and 10 speak Spanish.

(In reply to re(2): Solution (???) by nikki)

Nikki:

Tristan showed that the method of your proof is invalid by applying it to a different problem, and then finding a counter-example.  This doesn't mean a counterexample exists in the original problem--it just means that the proof is flawed.

Thinking about it, I think I can say more directly why it's flawed.  You have successfully shown how to find combinations of people from 50 to 150 that satisfy the problem, but there are certainly more than just those 101 combinations of people-languages such that 50 speak English, 50 speak French and 50 speak Spanish. 

In other words, your proof sucessfully shows that there is at least one combination of, say 87, people that satisfy the 50-English, 50-French and 50-Spanish requirement that can be divided up into 5 groups of 10-10-10, but it doesn't show that ALL POSSIBLE such combinations of 87 people can be divided up that way.  Does that make sense?

Posted by Ken Haley on 2004-11-18 05:52:38