Numbers for Words People

Fifty people enter a room. If each wants to shake the hand of every other person once, how many handshakes will take place? This problem is a classic of mathematics and we will solve it with the help of some visuals.

You are one of the fifty people. You need to shake hands with the other 49. Jim has already shaken your hand and needs to shake hands with the other 48. Sandra has shaken hands with you and Jim, so she has another 47 handshakes to go. This goes on until the second last person shakes hands with the last person. The latter has no need to initiate any handshakes.

Solving a smaller version of the problem

Mathematicians are too lazy to carry out such a long string of additions. They’d rather spend time working out a general solution. They would then apply their solution to any number of party attendees. To do this, they will start with a smaller number, like 10 people.

As promised earlier, we will take the visual route to the solution. In the triangle below, dots represent handshakes between 10 people. The top row shows the 9 handshakes that the first person needs to execute. The second row shows the second party goer’s 8 handshakes, and so on.

Wait! Don’t count the dots. We need to find a “clever” way to find the number of handshakes, which can then be applied to the original problem that has 50 people in the room.

Notice that the blue dots in the above triangle are half of the dots in the following, 10 by 9 triangle.

Since there are 9 rows of 10 dots, the total number of dots is $9 \times 10 = 90$ dots. As only half of these dots represent handshakes, the total number of handshakes$= 90 \div 2$, or 45 handshakes.

Back to the original problem

Back to having 50 people at the party. Given what we learnt from the smaller instance of the problem, we will have a triangle with 49 dots at the top, 48 in the second row, all the way to one dot in row 49 at the bottom. Adding maroon dots will give us 49 rows of 50, or 49 $\times$ 50 = 2450 dots.

Again, following what we did above, we halve that number and end up with 1225 handshakes. So, our answer is:

It takes 1,225 handshakes for 50 people in a room to shake each other’s hand
exactly once.

The general case

Some may be satisfied with the answer to the handshake problem when 50 people are involved. A mathematician asks the question: What if an arbitrary number, n, were at the party?

Step 1: First, imagine the right-angled rectangle in Figure 1. We will have (n – 1) dots in the top row. Remember that, when there were 10 people, we had 9, or (10 – 1) dots in the top row, representing the 9 handshakes that the first person needs to execute. We keep going till we get to 1 dot at the bottom of our triangle.

Step 2: Now, let’s reproduce Figure 2 in our minds. we end up with (n – 1) rows of n dots. This means that our rectangle of dots will have $(n-1) \times n$ dots in total.

Step 3: We now need to halve the result from step 2, which give us the general formula:

The horizontal line, in this context, stands for division by 2.

Conclusion

In this post, we looked at the classic “handshake problem”. More importantly, we thought like a mathematician by solving an instance of the problem that involved operations on small numbers. We then saw that the same approach would work on a larger, but still manageable instance of the problem. Finally, we generalised the solution so it would apply to all instances of our problem.

I have written elsewhere about the concept of triangular numbers. See: Dots, numbers and the mind of a child genius, part 1 and part 2.

I would be grateful to you, dear reader, if you would give me some feedback on this article, both in relation to its content and writing style. Leave me a comment to let me know whether you found the topic interesting and the style clear. I am particularly interested to learn whether the generalisation section was worthwhile.