Pigeonhole Principle - Show two subsets have the same age 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.

To expand on what Ross said, your holes will be the total age of the group. How many holes are there? This is equivalent to asking what the minimal and maximal possible total ages are. Your pigeons will be the groups (this is the number of subsets).
When you find these two numbers, if the number of pigeons is larger than the number of buckets, you know you have two groups (not necessarily disjoint) with the same total age.
