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?

Guest Dec 25, 2020

#1**+2 **

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. **

Melody Dec 26, 2020