10 people, 2 groups with equally summed ages 
In a room there are 10 people, none of them is younger than 1 or older than 100 years. Prove
  that among them, one can always find two groups of people (possibly intersecting, but different)
  the sums of whose ages are the same.
Hint: Use the pigeonhole principle. How many different groups of
  people can be picked? How large the sum of their ages can be?

I don't know where to begin. My professor tackled a similar question to this in class, but he used a diagram with a smaller amount of people (6) and did not solve it combinatorically. I also began this problem assuming it meant pairs of people, but from what I can see it can include a group of any size (excluding one..right? I mean a group of one wouldn't have a sum). 
If someone could help destruct this with me, that would be appreciated. 
 A: There are $2^{10} - 1 = 1023$ groups we can form (or one less if we discount the total group).
The maximal sum is $10 \cdot 100 = 1000$, the minimal one $1 \cdot 1$. So at most 1000 different sums.
A: Unpacking this a little more:
For a particular selection of group from the $10$ people, each person is either in that group or not, so you can specify any group by stating "in" or "out" for each person of the $10$. To generate all groups you multiply all those "in-out" binary decisions across all the people: $2^{10}=1024$. One of the results is the empty group, and one is the group of all $10$ people, which are not relevant to the problem at hand, so we can consider the space of $1022$ selected groups (ranging in size from $1$ to $9$ people).
Now the maximum possible group age total is $900$ and the minimum is $1$, giving $900$ different possible age totals, so by the pigeonhole principle there must be at least two possible groups with the same age total, which answers the given question.
We can push the result a little further. Note that there are only 10 groups with 9 members, but if we exclude them then the maximum possible total drops significantly: $1012$ possible groups across the range of age totals from $1$ to $800$. Similarly we can successively exclude groups of size $8$, $7$ and $6$. This leaves :
$$\sum_{k=1}^5 {10 \choose k}=637$$ 
possible groups of sizes $1$ to $5$ with $500$ as the maximum age total, so again the pigeonhole principle tells us that there will at least $2$ groups with the same age total. Furthermore we can drop the size $1$ and $2$ groups from consideration and still have enough possible ways of choosing groups of size $3$ to $5$ - $582$ - to ensure that two of those groups have the same age total.
