Pigeonhole Principle - Show two subsets have the same age

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?


Hint: how many subsets are there? What is the maximum total age of a subset?


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.

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Combinatorics

Pigeonhole Principle

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