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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?

Logic  Nov 2, 2018
 #1
avatar+3282 
+3

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.

tertre  Nov 3, 2018
 #2
avatar+91360 
+2

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

 

 

cool cool cool

CPhill  Nov 3, 2018
 #3
avatar+91360 
+1

Heck.....tertre beat me to it  !!!!!

 

 

 

cool cool cool

CPhill  Nov 3, 2018

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