Three couples go to the movie theater. If they sit in a row at random, what is the probability that at least one couple is sitting together?
+118608 The number of ways the 6 people can sit is 6! = 720 ways
Let the couples be AB, CD, and EF
The number of ways that AB can sit together is 5!*2 = 240 ways
The number of ways that CD can sit together is 5!*2 = 240 ways
The number of ways that EF can sit together is 5!*2 = 240 ways
The number is ways that they can all sit in pairs is 3!*2^3 = 48
I have already added this in 3 times so I will have to subtract it twice.
How many ways can there be in 2 pairs only.
The odd pair out will have to be in in the 3rd and 6th chairs
or the 1st and 6th
or the 1st and 4th
So I have 3 * 3C2 *2^3 = 72 ways
I have added each of these in twice so I have to subtract once
So I get that the number of ways for at least 1 couple to be stated together is
3*240 - 2*48 - 72 = 552 ways
So P(at least one couple sits together) = 552/720 = 23/30
If you get confirmation that this answer is either correct or incorrect, please let me know.
