yupp the answer is 2/19......
see the approach shud be like this.....
out of 20 ppl, A and B are not to be selected, thrfore no of ways of selecting 6 ppl out of rest 18, is 18C6 and arranging them in 6 places is 6!, i.e. 18P6
now no of ways of arranging the left 12 persons is 12!
so total no of ways of arranging 18 ppl between A and B such that 6 are on one side and 12 on other is 18P6 . 12!
moreover A and B can be mutually arranged in 2! ways ....
thrfore total no of ways of arranging all 20 ppl in desired way = 18P6 . 12! 2!
and total no ways of arrangin 20 ppl around a round table= 19!
so the required probablity = (18P6 . 12! 2!) / 19!
= 2/19
-----------------cheers------------------