Question

Twelve students sit around a circular table. Let three of the students be `A`, `B` and `C`. Find the probability that `A` does not sit next to either `B` or `C`?

Answer 1

Roughly `65.5%`

Explanation:

Let's say that there are 12 seats and number them 1 - 12.

Let's put A into seat 2. This means B and C can't sit in seats 1 or 3. But they can sit everywhere else.

Let's work with B first. There are 3 seats where B can't sit and so therefore B can sit in one of the remaining 9 seats.

For C, there are now 8 seats where C can sit (the three that are disallowed by sitting on or near A and the seat occupied by B).

The remaining 9 people can sit in any of the remaining 9 seats. We can express this as `9!`

Putting it all together, we have:

`9xx8xx9! = 26,127,360`

But we want the probability that B and C don't sit next to A. We'll have A stay in the same seat - seat number 2 - and have the remaining 11 people arrange themselves around A. This means there are `11! = 39,916,800` ways they can do that.

Therefore, the probability that neither B nor C sit next to A is:

`26127360/39916800=.6bar(54)~=65.5%`