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------------------