Fifteen persons, among whom are A and B, sit down at random at a round table. The probability that there are 4 persons between A and B is

none of these

Fifteen persons can be seated around a round table in 14! ways.

So, total number of ways = 14!

Four persons out of 13 persons (excluding A and B) can be selected in ${^{13}C}_4$ ways. Considering these four persons and A and B as one individual there are 10 persons who can sit at a round table in 9 !ways. But, 4 persons between A and B can be seated in 4! ways. Also, A and B can interchange their positions. So, total number of ways in which 4 persons can be seated between A and B is

${^{13}C}_4 × 9!× 4!× 2 = 2×13!$

Hence, required probability $=\frac{2×13!}{14!}=\frac{2}{14!}=\frac{1}{7}$