Question

Twenty persons among whom $A$ and $B$, sit at random around a round table, then the probability that there are any $6$ persons between $A$ and $B$ is

• A. $\frac{2}{19}$
• B. $\frac{17}{19}$
• C. $\frac{{}^{18}{C}_6\times 2!}{19!}$
• D. $\frac{{}^{18}{C}_6\times 2!\times 2!}{19!}$

Answer 1

The correct option is A $\frac{2}{19}$

The number of ways in which 18 people (excluding A and B) sit in a round table with 6 people between A and B = ${}_6^{18}C\ast 2!\ast (13-1)!$

Thus, probability = $\frac{{}_6^{18}C\ast 2!\ast (13-1)!\ast 6!}{(20-1)!}=\frac{18!\ast 2!\ast 12!}{12!\ast 6!\ast 19!}=\frac{2}{19}$

Here, ${}^{18}{C}_6$ refers to selection of $6$ people out of $18,2!$ is arranging A and B, $6!$ is arranging the 6 people in between A and B.