Question

10 gentlemen and 6 ladies are to sit for a dinner at a round table. The probability that no two ladies sit together is

• A. $\frac{9!\times 5!}{15!}$
• B. $\frac{9!{\times }^{10}{P}_6}{15!}$
• C. $\frac{9!{\times }^{10}{C}_6}{15!}$
• D. $\frac{9!{\times }^{10}{C}_5}{15!}$

Answer 1

Answer: (B)

$\frac{9!{\times }^{10}{P}_6}{15!}$

The total number of ways in which 16 people can sit around a round table is 15!
Now the number of ways in which 10 men can sit around round table is 9!
Now, there are 10 vacant positions between all consecutive men, we can arrange 6 women in ${}^{10}{P}_6$ ways
so the probability = $9!\ast {(}^{10}{P}_6)/15!$