Please help me find out where I made mistake. 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 four girls stand consecutively in the queue. Then the value of $m/n$ is?
$n$ is clearly $5!6!$
For $m$ I have considered 4 girls to be one unit. They can be arranged in $_5P_4$ ways. Now there are 5 boys, this 4-girls unit and an extra girl. So they can be arranged in $7!$ ways, so total $7!5!$ ways. However this also includes the cases where 5 girls come together, so I subtracted $5!6!$ from here, so $m=7!5!-6!5!$
Hence $m/n$ comes out to be 6, but the answer given is 5.
 A: Your value for $n = 5!*6!$ is correct.
For the next one, let's remember that there are 5 ways to pick a girl to be excluded. Then, using our group of 4 girls together and 5 boys, that will be $6!*4!$. 
Now, we can add our extra girl to get $6!*4!*5$ because of the seven gaps in the line, she can only join at 5 places (to avoid being next to the other girls). 
Finally, there were 5 ways to pick the extra girl, so we get $m = 6!*4!*5*5$. Now, dividing $m/n$, we get $\frac{6!*4!*5*5}{6!*5!}$ which gives us the desired answer of 5.
EDIT:
The flaw behind $7!*5! - 6!*5!$ is a little hidden. Assume we have girls $A, B, C, D, E$. Also, briefly assume that the "extra" is $E$. In $7!*5!$, we get some the possible organization structures for the girls, including $\underline{A,B,C,D}$ $\underline{E}$, and then the boys. Notice how no matter what, $E$ will never be in the "inside" of the girls. However, $6!*5!$ subtracts out parts that don't exist, such as $A, B, C, E, D$. Therefore, you receive the wrong answer.
A: You can consider  the $5$ consecutive girls as a block ($X$) to calculate $n$. We do not distinguish between the specific girls/boys. Then the elements  are
$b,b,b,b,b,X$
Then there are $n=6$ ways of arrangements.
Now you have a block ($Y$) with 4 consecutive girls and one lonely girl. The elements are 
$b,b,b,b,b,Y,g$
Without any condition there are $\frac{7!}{5!}$ ways of arrangements.
Now we have to substract the number of ways with the to blocks $gY$ and $Yg$
$$ \ \_ \ b \ \_ \ b \ \_ \ b \ \_ \ b \ \_ \ b \ \_ \ $$
The block $\boxed{Yg}$ can be placed in one of the $6$ gaps. Similar for $\boxed{gY}$. Thus $m=\frac{7!}{5!}-12=6\cdot 7-12=30$
A: Let $n$ be the number of ways that $5$ boys and $5$ girls can stand in a queue so that the $5$ girls stand together.  The $5$ girls can be arranged in $5!$ ways, and similarly the order in which the $5$ boys are placed from left to right can be chosen in $5!$ ways.  Once these two orders are chosen, we need to decide how many boys are to be placed to the left of the girls. This can be done in $6$ ways (since the number of boys to the left can be $0,1,\ldots,5$).  Hence $n=5! 5! 6$.
Let $m$ be the number of ways in which the $5$ boys and $5$ girls can stand in a queue so that exactly $4$ girls stand together. The, the arrangement has the form $-gggg-g-$ or $-g-gggg-$, where the dashes represent zero or more boys and such that the middle dash has at least $1$ boy.  The number of arrangements of the form $-gggg-g-$ is the number of ways to arrange the $5$ girls (which is $5!$) times the number of ways to arrange the $5$ boys from left to right (again $5!$) times the number of ways to place the $5$ boys in the $3$ dashes such that the middle dash has at least $1$ boy. After placing $1$ boy in the middle dash, the remaining $4$ boys need to be partitioned into $3$ parts, which can be done in ${6 \choose 2} = 15$ ways. Thus, the number of arrangements of the form $-gggg-g-$ is $5! 5! 15$. Similarly for the other kind of arrangement. 
Hence, $m=5! 5! 15 + 5! 5!15$.  So $m/n = 30/6 = 5$.
