SOLUTION: 7 people are seated around a table. What is the probability that Jack and Jill are not seated together.

For some reason, when problems are about sitting around a table,
all the rotations as shown below are considered the same.  (I don't 
believe they should be the same, but mathematicians do consider them
the same seating arrangement, so I will here.  You might want to ask
your teacher about whether that should be the case.)



I am assuming that all 7 of those above are considered as the
same seating arrangement.

If it doesn't matter where Jack and Jill sit, the number of ways is 7!/7.
(We divide by 7 because all the 7 seating arrangements
pictured above are considered the same.  And 7!/7 is just 6!.

Now we must calculate the number of ways Jack and Jill sit together.
Imagine there being only 6 chairs and Jill sits in Jack's lap.  That's 
the same as when Jack is seated left of Jill.  That's 6!/6 or 5! ways.
But there is another, by imagining only 6 chairs with Jack sitting in 
Jill's lap.  That's the same as when Jack is seated right of Jill. 
That's another 6!/6 or 5! ways. So there are 2*5! ways they sit together. 

P(they sit together) =  2*5!/6! = 2*(5*4*3*2*1)/(6*5*4*3*2*1) = 2/6 = 1/3

Therefore P(they do not sit together) = 1 = 1/3 = 2/3.

Edwin