Eleven boys and girls wait to take their seats in the same
row in a movie theater. There are exactly 11 seats in the row.
They decided that after the first person sits down, the next
person has to sit next to the first. The third sits next to
one of the first two and so on until all eleven are seated.
In other words, no person can take a seat that separates
him/her from at least one other person.
How many different ways can this be accomplished? Note that
the first person can choose any of the 11 seats.
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Answer Posted / puchi mukhujje (:-p)
i think you're trying to say that each person is previously numbered from 1 to 11.
the first person can choose his seat between the two corner seats and 11 other seats.if he chooses one of the corner seats which he can do in 2 ways,then because there is only one vacant seat beside him, the 2nd person will have to sit on that seat and so on.
now let us mark the middle seats from 1 to 9.now if he picks the nth seat,you notice,after they're all seated, that on both sides of the 1st person people are seated with their numbers in ascending order.when 1 chooses the nth seat, the no. of ways in which they can be seated becomes 10Cn. 1 can choose any seat from 1 to 9.
so the total no. of ways is 2^10-2+2=2^10 (ans.)
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