Suppose that there are $10$ people at a party whose (integer) ages range from $0$ to $100$.
Show that there are two distinct, but not necessarily disjoint, subset of people that have exactly the same total age.
So my thoughts so far are that you could split the ages $0-100$ $(101$ numbers$)$ into $11$ groups (pigeons) and then categorize these groups into the $10$ people (holes). However this seems backwards and not sound to me. Is there a different way to think about this problem?
Any help is appreciated.
Best Answer
The sum of ages cannot exceed $100 \times 10=1000$ $(1001$ cages, including $0 )$. The total ways to chose subsets of people are $2^{10}=1024$ $(1024$ pigeons$)$.
Can you see that coming?