Four married couples attend a party. Each person shakes hands with every other person, except their own spouse, exactly once. How many handshakes?
Andrew Mclaughlin 
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.
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?
$\endgroup$ 87 Answers
$\begingroup$$8$ people. Each experiences handshakes with $6$ people. There are $6\times 8=48$ experiences of handshakes. Each handshake is experienced by two people so there $48$ experiences means $48\div 2=24$ handshakes.
$\endgroup$ $\begingroup$Suppose the spouses were allowed to shake each other's hands. That would give you $\binom{8}{2} = 28$ handshakes. Since there are four couples, four of these handshakes are illegal. We can remove those to get the $24$ legal handshakes.
$\endgroup$ 3 $\begingroup$You may proceed as follows using combinations:
- Number of all possible handshakes among 8 people: $\color{blue}{\binom{8}{2}}$
- Number of pairs who do not shake hands: $\color{blue}{4}$
It follows:$$\mbox{number of hand shakes without pairs} = \color{blue}{\binom{8}{2}} - \color{blue}{4} = \frac{8\cdot 7}{2} - 4 = 24$$
$\endgroup$ $\begingroup$Let's look at it not from individuals, but from couples. There are four couples, i.e. $3!=6$ meetings of couples. Per meeting of couples, there are four handshakes. This makes it $6\times4=24$ handshakes.
Thanks @CJ Dennis for pointing out an error in the reasoning: It should, of course, be the sum, not the product, so the correct number of meetings of couples is$\sum_{k=1}^{n-1}k=\frac{n(n-1)}{2}$.
$\endgroup$ 0 $\begingroup$Each line is a handshake between the required two people. There are 24 lines:
$k$ couples entails $2k$ people. If we imagine each couple going in sequential order, couple 1 will each have to shake $2k-2$ couple's hands for each individual, or $4k-4$ handshakes for couple 1 total. Since there is 1 fewer couple every time a new couple shakes hands, there will be $4k-4i$ handshakes by the $i$-th couple. So the total number of handshakes is given by:
$$\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.
$\endgroup$ 1 $\begingroup$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
$\endgroup$ 3
