0
$\begingroup$

I'm attempting this question, but I'm a little unsure about my answers. The full question is:

Five people are to be seated around a circular table. Two seatings are considered the same if one is a rotation of the other.

  1. How many seatings are possible?
  2. How many are possible if John and Mary always insist on sitting next to each other?
  3. What is the probability if the five people are place randomly around the table that John and Mary will end up sitting next to one another?

My answers are:

  1. (5-1)! = 4! = 24
  2. 3! = 6
  3. C(5,2) = 10
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

For b, John and Mary can sit beside each other two different ways. For c, a probability has to be less than or equal to $1$, so you have to divide by the number of possibilities.

$\endgroup$
2
  • $\begingroup$ Just making sure, the answer for b would be 12; and c would be (12/24) * 100 = 50%? $\endgroup$
    – user754950
    Dec 15, 2011 at 14:14
  • $\begingroup$ @user754950: right. For c, if you seat John first, then Mary has two choices next to him and two that are not, an easy check. $\endgroup$
    – Ross Millikan
    Dec 15, 2011 at 14:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .