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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?

 Dec 25, 2020
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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. 

 Dec 26, 2020

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