Three couples go to the movie theater. They want to seat people together for maximum enjoyment, but instead they randomly file into a row with six seats. What is the probability that they sit in a socially optimal configuration, in which each person is sitting next to his or her partner?
Hmm, not that good at these type of problems, but I'll give a shot at it!
There are 3! or 6 ways to choose the three couples to sit.
So, 6*8=48 ways to sit three couples, and 8 ways to choose the order the couples sit in.
Finally, there are a total of 6!=720 ways the six individual people can sit.
Thus, 1/15 is our answer..simplified from 48/720.
I think this might be correct.....
Call the couples AB CD EF
Each couple has 2 ways that they could sit together [ example, AB or BA ]
And we have 3! ways to seat the couples themselves
So....this gives 2^3 * 3! = 48 ways that they could sit together
And the total possible arrangements are 6! = 720
So the proability of a socially optimum configuration is just 48 / 720 = 1 / 15