One of the problems with the "round table" seating is in the interpretation.
The most common one is that if everybody stood up and moved one seat to the right, the seating arrangement is still the same.
So consider laura and Adrian seated. Then there are 10! ways to sit everybody else going clockwise and 10! counterclockwise. Which would be 2*10! ways.
But Laura and Adrian could also be switched
My answer : 2*2*10! or 4*10! ways to seat with Laura and Adrian together
without restrictions, there would be 2*11! ways , sit the first person, then the other 11 either clockwise or counter-clockwise
prob of the event = 4*10!/2*11! = 2/11
Laura and Adrian attend a dinner party. The 12 guests are seated at random
around a round table, with all seating arrangements equally likely.
(a) What is the probability that Laura and Adrian sit next to each other.
(b) Explain your answer by giving a clear description of an equally-likely-
outcomes model on which it is based. In other words, tell me what
you are using, how many elements it has, and how this implies your answer from
part (a).
why is the answer 2/11 in part a and not 2/12? Thanks
1 answer
1 answer