Question

Four men and three women are standing in a line for railway ticket. The probability of standing them in alternate manner is

• A. $\frac{1}{35}$
• B. $\frac{1}{33}$
• C. $\frac{1}{84}$
• D. $\frac{1}{7}$

Answer 1

The correct option is A $\frac{1}{35}$

Total number of ways of standing four men and three women in a line
$=7!$
Number of ways of standing them in alternate manner
$i.e.\left\{MWMWMWM\right\}=4!\cdot 3!$
Required probability
$=\frac{4!\cdot 3!}{7!}=\frac{6}{7\cdot 6\cdot 5}=\frac{1}{35}$