Four couples go to a party. They decide to split randomly into groups of 2. What is the probability that no group has a couple that came together.
I actually have an MS in math; feel free to use mathematics at any level. The answer is supposed to be 37.5% or 1-37.5 I have forgotten.
The Attempt at a Solution
I've tried several different approaches.
First: Total ways possible is 8! / 2!2!2!2! OR you can do: 8C2 * 6C2 * 4C2 * 2C2 (which is the same thing); I believe this is correct.
Total ways of getting no couples (this part is incorrect). 7P6 * 5P4 * 3P2 * 1P1 = 7*5*3*1.
Second: There are 8 people, label them 1-8. 1,2 = couple, 3,4 = couple, 5,6 = couple, 7,8 = couple. This gives us 28 different combinations of couples that are possible. 7+6+5+4+3+2+1(because 1 can be paired with 2-8 = 7 ways, 2 can be paired with 3-8 = 6 etc).
But I can't figure out all the different cases; mostly because I don't want to do it this way. I feel like there should be an easier way to solve this that I'm over looking. The book solves this problem by setting up an experiment.