Ok, this should be an easy one for anybody who is not me … or obviously anybody else in my class. Since nobody got it, the Prof gave us the weekend to work on it.
No fret answering: It was in-class work but he said there will be at least one similar question on an upcoming quiz. I just need to know how the heck to do it.
…You are at a party. One hundred people are present (99 plus you). If everyone shakes hands exactly once with everyone else, how many handshakes will occur?
Help?
You shake hands with everybody. That’s 99 handshakes. Your friend shakes hands with everybody. Not counting him shaking hands with you (Because then we would be counting that handshake twice) that’s 98 handshakes. So the answer will be 99+98+… I think you can take it from there.
Ah, combinatorics! My most-favorite of all maths!
Too bad I’m not better at it. 
At any rate, there’s a specific name for this type of solution, and even a single key to do it on some calculators… wink
99+98+97+96…
=4950
The general formula for n+1 people would be (n^2+n)/2
Every pair of people shakes hands once, so you just need to know how many pairs there are. The answer is the number of ways to choose two people from a group of 100. In general, there are n!/(k!(n - k)!) ways to choose k people from a group of n, and when k = 2, that reduces to n(n - 1)/2.
Note that we’re counting Bob and Joe and Joe and Bob as the same pair here; order is irrelevant. Also, in case you’ve forgotten, ! is defined as 0! = 1 and (n + 1)! = (n + 1)n!.
Perhaps the simplest way of looking at it is:
Each person shakes hands 99 times - sounds like 9900 handshakes
But each handshake involves two people, so it’s actually half that many (4950)
Another way of looking at it, setting up the problem based on the logic above, is this:
99+1=100
98+2=100
97+3=100
…
51+49=100
So, so far we have 4900. Add the 50 that’s left over, and we get 4950.
This is sometimes denoted as C(100, 2) or as [sub]100[/sub]C[sub]2[/sub] (which may help you find that calculator key that Sofaspud alluded to).
Yet another way to look at it: Imagine you are the first person at the party (like, you’re the host).
The first person to arrive shakes hands with you.
The second person to arrive shakes hands with you and person #1. There have now been 1+2 handshakes.
The third person to arrive shakes hands with the three people who are already there (you, person #1 to arrive, and person #2). With these three handshakes added in, there have now been a total of 1+2+3 handshakes.
And so on. After n people (not counting yourself) have arrived, there will have been 1+2+3+…+n handshakes total. (In the OP, n = 99.)
When you add up the whole numbers from 1 to n, the sum is n(n+1)/2, for reasons that are nicely explained here.
Thus, the answer to the OP’s question is 99(100)/2 = 4950.
FanTAStic. Thanks for all the suggestions, folks. I’ll write some of these down and share them with the kids in my group.