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?
Best Answer
$8$ 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.