Question

In a class of $10$ students, there are $4$ girls, number of different ways can they be arranged in a row such that no two of the $4$ girls are consecutive is

• A. $35(6!)$
• B. $30(6!)$
• C. $30(7!)$
• D. $6!$

Answer 1

The correct option is C $30(7!)$

Number of girls $=4$
Number of boys $=10–4=6$
X B X B X B X B X B X B X
$6$ boys an be arranged in a row in $6!$ ways. If girls are arranged at the places indicated by crosses, no two girls will be consecutive.
Required number of ways $={}^7{P}_3\times 6!$
$=\frac{7!}{4!}\times 6!=7\times 6\times 5\times 6!=30.(7!)$