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

Question

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

My book gave the answer as $24$. I do not understand why.

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I thought of it like this:

You have four pairs of couples, so you can think of it as

M1W2, M2W2, M3W3, M4W4,

where M is a man and W is a woman. M1 has to shake 6 other hands, excluding his wife. You have to do this 4 times for the other men, so you have $4\times 6$ handshakes, but in my answer, you are double counting.

How do I approach this problem?

Answer 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.

Answer 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.

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

Answer 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.

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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 $\sum_{k=1}^{n-1}k=\frac{n(n-1)}{2}$.

Answer 5

Each line is a handshake between the required two people. There are 24 lines:

Answer 6

$k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:

$$\sum_{i=1}^k (4k-4i) = \sum_{i=1}^k4k - \sum_{i=1}^k4i = 4k^2 - 4\frac{k(k+1)}{2} = 4(k^2 - \frac{k^2+k}{2}) = 4(k^2 - (\frac{k^2}{2} + \frac{k}{2})) = 4(\frac{k^2}{2}-\frac{k}{2}) = 2(k^2-k)$$

for $k$ couples. Plugging in $k$ = 4 verifies a solution of 24 for this case.

Answer 7

A simple approach:

There are 8 person in total.

Each one will shake hands with 6 others.

Total shakehands from individual perspective: 6*8 gives 48

Actual shakehands: 48/2 = 24