###### Brian McCall

Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school

Brian was a geometry teacher through the Teach for America program and started the geometry program at his school

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# Number of Handshakes at a Party - Problem 1

Brian McCall
###### Brian McCall

Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school

Brian was a geometry teacher through the Teach for America program and started the geometry program at his school

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The formula for the number of handshakes possible at a party with n people is

# handshakes = n*(n - 1)/2.

This is because each of the n people can shake hands with n - 1 people (they would not shake their own hand), and the handshake between two people is not counted twice.

This formula can be used for any number of people. For example, with a party of 10 people, find the number of handshakes possible.

# handshakes = 10*(10 - 1)/2.

# handshakes = 10*(9)/2.

# handshakes = 90/2.

# handshakes = 45

So, there are 45 handshakes that can be made between 10 people.

Let’s say you invited 10 people over for a party, how many possible handshakes would there be? And again I’m going to count a handshake as if you have two people meeting each other that would be one handshake. What we’re going to do is we’re going to use mathematical modelling which means you’re kind of using a picture to describe a problem.

So let’s say you had two people, a mathematically model of that would just be a line segment and you’d say that the number of handshakes possible here is one.

Okay so our goal is eventually to figure out for n number of people how many handshakes, so let’s look at a couple of more examples. Let’s say you had three people, the mathematical model there would be 1, 2, 3 people and there would be three handshakes, so three people three handshakes. You have four people you’re going to have four dots which represent for people at that party and you’re going to have four handshakes but you’re also going to have two more.

So we’ve got one, three a total of six. So I’m noticing that we don’t have a linear function here, we’ll do one last one. If you have five people at a party, you’re going to have five people and then you’re going to have five more handshakes for a total of 10.

So a couple of ways that you could do this, you could say oh well these are the triangular numbers so I know the formula or you could say well looking at my model how can I come up with the number of handshakes. Well the number of people we’re going to say is n and if I have one person. how many times could I shake hands. Here I could shake hands one time, which is one less than two. Here I could shake two times which is one less than three. Here I could shake one, two, three times. So I’m seeing that the number of handshakes is one less than the total number of people because I can shake hands with everyone there except for myself.

Just like with the diagonal problem we’re going to count every single one of these handshakes. So I’m going have to have divide that whole term by 2. So the number of handshakes at a party is the number of people times the number of people minus 1, because you’re taking yourself out of that equation and you’re going to divide it by 2 because you don’t want to double count every person, excuse me you don’t want to double count every handshake.

So that’s our formula for the number of handshakes with n number of people.