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Number of Handshakes at a Party - Concept

10,154 views
Teacher/Instructor 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

Mathematical modeling is used to generate a formula for the number of handshakes possible. The situation essentially asks how many sides and diagonals are in a given polygon and generates the triangle numbers: 1, 3, 6, 10, 15,

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 of two people meeting each other that would be one handshake. What we're going to do is we're going to use Mathematical modeling which means you're kind of using a picture to describe a problem. So let's say you had two people, the Mathematical 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 lets look at a couple more examples, let's say you had three people, the Mathematical model there will be 1, 2, 3 people and there will be three handshakes so three people three handshakes. If you're four people, you're going to have four dots which represent the four 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 3 a total of 6 so I'm noticing that we don't have a linear function here. We'll do one last one.
If you have 5 people at a party, you're going to have 5 people and then you're going to have 5 more handshakes for a total of 10. So a couple 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 a number of handshakes?" Well the number of people, we're going to say is n and if I am 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 1, 2, 3 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 from myself.
Just like with the diagonal problem we're going to double count every single one of these handshakes so I'm going to have to divide that whole term by two. So the number of handshakes at a party is the number of people times the number of people minus 1 cause you're taking yourself out of that equation and you're going to divide it by two 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.

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