Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes?

Solution 1:

$8$ people. Each experiences handshakes with $6$ people. There are $6\times 8=48$ experiences of handshakes. Each handshake is experienced by two people so there $48$ experiences means $48\div 2=24$ handshakes.

Solution 2:

Suppose the spouses were allowed to shake each other's hands. That would give you $\binom{8}{2} = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.

Solution 3:

You may proceed as follows using combinations:

• Number of all possible handshakes among 8 people: $\color{blue}{\binom{8}{2}}$
• Number of pairs who do not shake hands: $\color{blue}{4}$

It follows: $$\mbox{number of hand shakes without pairs} = \color{blue}{\binom{8}{2}} - \color{blue}{4} = \frac{8\cdot 7}{2} - 4 = 24$$

Solution 4:

Let's look at it not from individuals, but from couples. There are four couples, i.e. $3!=6$ meetings of couples. Per meeting of couples, there are four handshakes. This makes it $6\times4=24$ handshakes.

Thanks @CJ Dennis for pointing out an error in the reasoning: It should, of course, be the sum, not the product, so the correct number of meetings of couples is $\sum_{k=1}^{n-1}k=\frac{n(n-1)}{2}$.