Hello, kmring!
If there are 100 people and each person shakes hands only one time with each person,
how many handshakes would that be?
The answer is 4950. My question is how do I put this into an algebra formula?
Assuming you're new to this type of problem, here's a walk-through.
Take any person in the group, say, person \(\displaystyle A\).
He can shake hands with any of the other 99 people . . . so he creates 99 handshakes.
This is true for each of the 100 people, so there are: \(\displaystyle \,100\,\times\,99\:=\:9900\) handshakes.
But this list includes entries like: "A shakes hands with B" and "B shakes hands with A",
\(\displaystyle \;\;\)which represent the same handshake. \(\displaystyle \;\)So our list is
twice as long as it should be.
Therefore, there are: \(\displaystyle \,\frac{100\,\times\,99}{2} \:=\:4950\) handshakes.
Generalization: For \(\displaystyle n\) people, there will be: \(\displaystyle \.\frac{n(n\,-\,1)}{2}\) handshakes.
You probably haven't learned Permutations and Combinations yet.
This is a Combination problem, written: \(\displaystyle \,_{_{100}}C_{_2}\,\) or \(\displaystyle \,C(100,2)\,\) or \(\displaystyle \frac{100!}{2!\cdot98!}\)