Numbers for Words People

We have already seen the classical form of the handshake problem, where we asked: If there were 10 people at a party and every person were to shake hands with every other person exactly once, how many handshakes would there be in total? In this post, we will solve the problem but with people arriving at the party as couples.

Defining our problem

If there were 10 couples at a party, and every person were to shake hands with every other person, except for their partner, exactly once, how many handshakes would there be in total?

As we did when we solved the classical form of the problem, we will answer our question for:

• Smaller instances of this problem, namely for 2, 3 and 4 couples;
• 10 couples;
• Any number of couples.

Solving a smaller problem

Figure 1 below shows us 2 and 3 couples, respectively, with lines (edges) showing the handshakes between them. Note that when a couple arrives at the party, they do not shake hands with each other, hence the broken edge connecting the couple.

When a fourth couple joins the party, they need to shake hands with each of the 6 people who make up the previous three couples (See figure 1.2). That’s an additional 12 handshakes.

To better remember our numbers, let’s compile them in a table.

Number of couples	Number of people	Number of handshakes
2	4	4
3	6	12
4	8	24

Table 1: Number of handshakes between 2, 3 and 4 couples, respectively.

Answering our question

Our question was about 10 couples. Let’s extend table 1 above:

• A fifth couple enters. That’s two people who need to shake hands with 8 existing attendees => Add 16 to the number of handshakes
• A sixth couple enters. That’s two people who need to shake hands with 10 existing attendees => Add 20 to the number of handshakes

Number of couples	Number of people	Number of handshakes
2	4	4
3	6	12 (previous + 8)
4	8	24 (previous + 12)
5	10	40 (previous + 16)
6	12	60 (previous +20)

Table 1: Number of handshakes between 2 to 6 couples, respectively.

To get to 10 couples, we can follow the pattern of additions in the last column:

Generalising our solution

As noted in prior posts, a mathematician would want to solve such a problem for a generic number of people. The formulation would be:

If n people were at a party, and if they were there as couples, how many handshakes would it take for each person to shake hands with every other person, except for their own partner?

We can answer this question by:

• Working out the classical form of the problem, for n people, without the parter exception; then
• Subtract the number of couples, which is half of the number of people, from the total

We have already seen that n people would require $\frac{(n-1) \times n}{2}$ handshakes. We simply have to “break the link” between the couples. That’s one handshake for each of $\frac{n}{2}$ couples that needs to be subtracted.

The calculation for the number of handshakes among n people, provided no one shakes hand with their own parter is:

$$\frac{(n-1) \times n}{2} – \frac{n}{2}$$

We can further simplify this mathematical expression, but this is beyond the scope of our post.

Over to you

Why don’t you try to do your own calculation to find the number of handshakes for 12 couples. You can either use the generalised expression or simply extend the table in the section entitled Answering our question. Let me know your result in the comments.