# 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?

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%