[Feature photo above by Tobias Wolter (CC-BY-SA-3.0) via Wikimedia Commons.]
If seven people meet at a party, and each person shakes the hand of everyone else exactly once, how many handshakes are there in all?
In general, if n people meet and shake hands all around, how many handshakes will there be?
Our homeschool co-op held an end-of-semester assembly. Each class was supposed to demonstrate something they had learned. I threatened to hand out a ten question pop quiz on integer arithmetic, but instead my pre-algebra students presented this skit. You may adjust the script to fit the available number of players.
7 friends (non-speaking parts, adjust to fit your group)
Each friend will need a sheet of paper with a number written on it big and bold enough to be read by the audience. The numbers needed are 0, 1, 2, 3, … up to one less than the number of friends. Each friend keeps his paper in a pocket until needed.
[As the first narrator begins to speak, the friends come on stage and silently greet each other.]
Narrator #1: Seven friends meet each other in the hallway before class. Being polite young ladies and gentlemen, they want to shake hands with each other, and they don’t want to hurt anyone’s feelings or leave anyone out.
[Friends may give the narrator a strange look, but then they begin to shake hands all around.]
Narrator #1: If each friend shakes the hand of every one of the other friends exactly once, how many handshakes will there be?
[Narrator #1 takes guesses from the audience, which will almost surely be much larger than the actual answer. For best results, avoid taking guesses from any audience members who look like they are actually doing mental arithmetic. For instance, I refused to call on anyone from the MathCounts team.]
[Meanwhile, the friends on stage pantomime counting each other, counting on their fingers, and generally looking puzzled. As Narrator #2 begins speaking, the friends move to the side of the stage.]
Narrator #2: In our math class, we learned a way to organize this problem so that we can find the answer. We start with small numbers and work our way up. If there is only one person, there can’t be any handshakes.
[Friend with the “0” comes back to center stage, tries to shake hands with himself, shrugs, and then holds up his paper. He will continue to hold the “0” where it is visible for the rest of the skit.]
Narrator #2: If another friend comes along, there is one handshake.
[Friend “1” comes out, shakes hands with “0”, and holds up his paper. As the narrator continues, each friend in turn comes out and shakes the hands of everyone in turn, then holds up the number on his paper.]
Narrator #2: When the next friend comes along and shakes both their hands, that adds two more handshakes to the party…
The next friend adds three handshakes to the total…
The next friend adds four more…
The next friend adds five…
And the last friend shakes all six of their hands.
Narrator #3: Now, how can we add up all those handshakes to find the total number? The easiest way to add a long list of numbers is to rearrange them into tens.
[Friends on stage rearrange themselves into small groups as Narrator #3 continues.]
Narrator #3: We can see that 6 and 4 together are 10. Then 2, 3, and 5 make another ten. That is 20 so far, and 0 and 1 bring our total up to 21 handshakes in all.
[All players bow to the audience. Exeunt.]
We did not extend the skit to teach the general formula, but it is an interesting problem for beginning algebra students: If n people meet and shake hands all around, how many handshakes will there be? Can you figure it out?
11 thoughts on “Skit: The Handshake Problem”
My husband and I think it’s 45 handshakes
Two ways to think about it:
As in the skit, we start with zero and add up to ONE LESS than the number of people in the group. 0 + 1 + 2 + 3 + … + (n-1) = (n)(n-1)/2
(Incidentally, the induction proof for this sum formula is really fun for bright middle schoolers — with supervision.)
Each person (n people) shakes hand with everyone else (n-1 people), making a total of (n)(n-1) handshakes. But since I counted shaking hands with you and you counted shaking hands with me too, we’ve counted every handshake twice, so we have to divide our result by 2.
The formula is also the expression for C(n,2), but that’s probably too advanced for the audience you’re talking about! This a great problem to show how fundamentally different methods arrive at the same result.
Another way of visualising the formula
1 + 2 + 3 + … + n = 1/2 n (n – 1)
is that the numbers form a triangle. Suppose each number is a collection of stones (which I will illustrate with the letter ‘o’), such that 1 is a single stone, 2 is a pair of stones and so on. You could express the summation 1 + 2 + 3 as the following:
o o o
To compute the summation, count the total number of stones, which is clearly 6 here. Hence 1 + 2 + 3 = 6.
Now, imagining the general case, you would have a triangle as follows:
o o o
o o o o o o … o
The triangle has n rows and the last row has n stones in it. We want a way of finding out the number of stones in this triangle. Well, if we took an exact copy, turned it upside down and placed it alongside our original triangle, we’d form a rectangle. Let’s illustrate that in the case of n = 5:
o o o o o o
o o o o o o
o o o o o o
o o o o o o
o o o o o o
(The stones in the copied triangle are in boldface to make things clearer.) Now, we have a rectangle here, which still has n rows, but each row has n + 1 stones in it. We can easily count the number of stones in a rectangle by multipying these dimensions together: the number of stones in the rectangle is n (n + 1). But as the rectangle was formed by two identical copies of our original triangle, the number of stones in our original triangle must be 1/2 n (n + 1).
Oops, typo. Of course, that first formula should have read:
1 + 2 + 3 + … + n = 1/2 n (n + 1)
Being a visual thinker, I’ve always liked that way of showing the triangular number formula. It sticks in my mind, while the formula itself would have just fallen through the sieve. And over the past year of teaching MathCounts problems, I was surprised at how useful the triangular numbers were for many situations.
David, you’re right that the triangular number formula is:
But because one person cannot do a real handshake with himself, the handshake formula is, as the ladies said:
A more mechanical interpretation of the solution:
For a handshake, two different persons should be chosen from N people.
First person can be chosen in N different ways.
Now there are only (N-1) choices for the second person to be chosen.
So total number of ways of chosing 2 different people for a handshake would be N(N-1)
For a handshake, since the order of chosing people is not important, there are duplicate handshakes. Every handshake is repeated twice in N(N-1) handshakes and hence N(N-1)/2 effective handshakes.
Venkat, when we did the problem as class homework, that was our preferred solution. Short and easy! But when we put the problem into a skit for our end-of-session assembly, the add-it-all-up solution was easier to act out.
When any number of people meet each other how many handhsakes are needed so that everyone has greeted everyone only once?
And what is the rule for ANY number?
Look in the comments. Judy, Sara, and Venkat all explained it in different ways, so you can choose the explanation that makes the most sense to you. Whatever number you want to use, that is the n in the formula. David was talking about a related topic: Triangular numbers.
well, very good skits n it can increase the mental leval of students if it can be done in assembly.