standing in a queue is easy, isn't it

I know the solution is long but It is worth reading (I feel so). Plz read it completely and tell me if there is a problem.

We are required to find $\cfrac mn$

Case 1: to find $n$

There are 7 women. So let us consider all 7 together as 1. Then the total number of people standing in the queue would be 6 ( 5 men + 1). They would arrange themselves with $6!$ .And the 7 ladies would arrange themselves as $7!$

By multiplication rule

$n = 6! \times 7!$

Case 2: to find $m$

First we need to select 6 women out of 7. It happens by $\dbinom 76$ . These 6 women arrange in $6!$ ways.Thus total is $7!$ . Then , let us consider these 6 women as $x$ , the 7th woman as $y$ . Clearly there are 7 people in the line to stand $x+y+5$ .

$m =$ total number of ways to stand for 7 people - the situations in which x and y stand next to each other ( when x and y stand next to each other the case of m is violated)

Total number of ways $= 7! \times 7!$

$\huge \text{\_ \_ \_ \_ \_ \_ \_}$ Consider these 7 dashes for the 7 people to stand. There are $6 \times 2$ ways for x and y to stand together. The 5 men arrange themselves as $5!$

Situation in which m case is violated $= 7! \times 12 \times 5!$

Thus $m = (7! \times 7!)-(7! \times 12 \times 5!)$

Therefore $\dfrac mn = \dfrac {(7! \times 7!)-(7! \times 12 \times 5!)}{7! \times 6!}$ $\implies \boxed {\dfrac mn = 5}$