Question

If there are $6$ girls and $5$ boys who sit in a row, then the probability that no two boys sit together is :

• A. $\frac{6!6!}{2!11!}$
• B. $\frac{7!5!}{2!11!}$
• C. $\frac{6!7!}{2!11!}$
• D. none of these

Answer 1

The correct option is A $\frac{6!6!}{2!11!}$

Q. If there are 6 girls & 5 boys who sit in a row,

then the probability that no two sit together

Sol. Total no. of arrangement can be $\to$ 11! ways

[for 6 girls & 5 boys]

Total no. of boys can be filled is $6P_5=6!$

Total no. of girls can be filled is $6!$

$\therefore$ So total no. of arrangements = $6!\times 6!$

$\therefore$ Probability is $\frac{6!\times 6!}{11!}$ i.e., Opt A