Question

The number of ways in which $6$ boys and $7$ girls can sit in a row such that they sit alternatively is

• A. $6!7!$
• B. $7!{}^8{P}_6$
• C. $2(7!{)}^2$
• D. $2(6!{)}^2$

Answer 1

The correct option is A $6!7!$

Solution:- (A) $6!7!$

Total no. of boys $=6$

Total no. of girls $=7$

$\therefore$ No. of ways if they sit together $=6!\times 7!$

Hence the correct answer is $6!\times 7!$.