please help me find why I'm wrong. How many ways to order boys and girls subject to restrictions 
Let $n$ be the number of ways in which $5$ boys and $5$ girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let $m$ be the number of ways in which $5$ boys and $5$ girls can stand in a queue in such a way that exactly $4$ girls stand consecutively in the queue. Then the value of $\frac m n$ is?

I first calculated $n= 5!6!$, no problem with that. BUT THEN for $m$ I first selected $4$ girls out of $5$ and arranged them. then considered them as a set and arranged the set along with the boys and the one lonely girl and subtracted the combinations for which all girls are together. Like this: $m=7!5!−6!5!$. Therefore $\frac m n$ equal to $6$. BUT THE ANSWER IS FIVE. I got the "gap" method used to calculate the answer. But I don't understand why my approach is wrong!
 A: How did you try to get $m$? First you ordered the boys: $5!$ ways. Then you selected the four girls $\binom{5}{4}=5$, ordered them ($4!$ ways), then placed them among the boys ($6$ ways), then placed the fifth girl among the boys.... presumably you said $7$ ways, five locations where the other girls aren't, then either first or last if you put them with the other girls? And then you took this $5!5(4!)(6)(7) = 5!7!$ and subtracted the ones in which all girls are together...
But your process miscounts $m$: consider the case where you line up girls $1$ through $5$ first, then boys $1$ through $5$. You counted this twice:

*

*Once when you left out girl $1$, and then decided to put her before the rest of the girls; and

*Once when you left out girl $5$, and then decided to put her after the rest of the girls.

So that's where the error lies in your computation.  Your $m$ is too big.
The double-counting occurs every time you have all girls together. So you actually need to subtract $n$ twice, not just once. Then you would get
$$5!7!-2(5!6!) = 5!6!(7-2) = 5!6!5.$$

If you don't want to worry about the double counting, you can do this:
Order the five boys: $5!$ ways. Now order the five girls: $5!$. Now select where the girls stand: there are six locations: first, after the first boy, after the second boy, and so on until they are after the fifth boy.
So $n = 5!5!6$, (same as you claim, so that part is right).
Now, for $m$; $5!$ ways to order the boys; $5$ ways to select the lone girl, and $6$ places where to put her. $4!$ ways of ordering the remaining girls, and $5$ places left where to put them (can't put them next to the lone girl). So $m=(5!)5(6)(4!)5 = 5!5!(6)(5)$ (same as we got before).
So
$$\frac{m}{n}=\frac{5!5!(6)(5)}{5!5!6} = 5.$$
