Question

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?

Answer 1

As there are 3 couples, so that means there are total 3 × 2 = 6 people altogether.

So now to make 6 people sit on 6 sits can be done in folowing ways:

suppose these are 6 seats available

_____1__ ______2___ ____3______ ___4__ ____5___ ___6____

Now there are 6 ways to make a person sit on seat 1.

Once seat 1 is occupied we are left with 5 people (1 person sat on seat 1)

so now 2nd seat can be filled in 5 ways

after 1 person sat on seat 2, then 3rd seat can be filled in 4 ways ( out of 6 people already 2 people sat on seat 1 and 2 so only 4 people left for 3rd seat)

And similarly for 4th seat can be filled in 3 ways

then 5th seat can be filled in 5 ways

then 6th seat can be filled in only 1 way.

Now as per fundamental principle of counting if there are m ways to do one event and n ways to do another event so altogether are m × n ways to do them.

So using same fundamental principle of counting here we get total ways to make 6 people sit on 6 seats as 6 × 5 × 4 × 3 × 2 × 1 or 6! = 720 ------------(1)

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Now they want each couple to sit together. So every two particular people make one group (1 couple).

so like we saw 6 people can be seated on 6 seats in 6! ways (from (1))

similarly 3 couples can be seated in 3! or 3 × 2 × 1 = 6 ways

And 1st couple can sit in 2 ways amoung themselves (husband left wife right or wife left and husband right).

similarly 2nd couple can also sit in 2 ways amoung themselves

and 3rd couple can also sit in 2 ways amoung themselves

So altogether, by funadmental principle of counting, all 3 couples can be seated in 6 × 2 × 2 × 2 = 48 ways -----------(2)

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Probability of event is given by formula:

$Probability = \frac{Number- of- favourable-outcomes}{Total-number-of -outcomes}$

so favourable outcomes from (2) we have as 48

total number of outcomes from (1) we have 720.

so plug these value in probability formula

Probability = $\frac{48}{720}$

To reduce this fraction divide by greatest common factor in 48 and 720 which is 48. so we get

$\frac{48 \div 48}{720 \ div 48} = \frac{1}{15}$

so thats the final answer.