Number of ways in which $5$ boys and $5$ girls are ordered in such a way that exactly $4$ girls stand consecutively in the queue 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 in such a way that exactly $4$ girls stand consecutively in the queue. Then find $\frac{m}{n}$.

I could find $n$ correctly as $6!\times 5!$, but I am finding $m$ as $7!\times 4!\times \binom{5}{1}$, but the book says $m$ is $6!\times 4!\times 5\times \binom{5}{4}$. Where have I gone wrong? What is the proper logic to find $m$?
 A: The logic of $7!\times4!\times\binom{5}{1}$ if clearly flawed. It also contains instances in which 5 girls were standing consecutively which is in contradiction with the demand of the question.
So, we can't do it the way you did. We have to make sure that $5$ girls do not stand consecutively in the queue.
The first part of the task is to identify a girl which will not be in the consecutive part of the queue. This can be done in $\binom{5}{1}$ ways.
Remaining $4$ girls will be standing consecutively but they may be standing in any order. So the number of orders possible for these girls is $4!$.
We have a group of $7$ objects, $4$ girls(counted as $1$), $1$ girl and $5$ different boys. These can be permuted in $7!$ ways but we have a restriction and we have to take care of that. First of all, we'll keep $4$ girls and $1$ girl in position with the given restriction. These $2$ objects can be kept in $7$ slots in $\binom{7}{2}\times2$ ways but we have $6\times2$ positions where we don't want these two groups to sit. SO the total no. of ways in which these $2$ groups can sit is:- $\binom{7}{2}\times2 - 6\times2 = 6\times5$.
Now the remaining 5 boys can sit in 5 slots in $5!$ ways.
Hence the value of $m$ is $\binom{5}{1}\times4!\times6\times5\times5!$. This explains the answer you found in the book.
A: *

*You can line up the boys in $5!$ ways.

*Next choose one of the $6$ gaps for the $4$ girls who are together.

*Then choose the $4$ girls and arrange them in order, which can be done in $\binom{5}{4}4!=5!$ ways.

*Finally, choose one of the $5$ gaps left for the remaining girl.


This gives $5!\cdot6\cdot5!\cdot5=5!\cdot6!\cdot5=432,000$ possibilities.
A: Your $n$ is correct. But about $m$ I have some thoughts.
First, we can remove a girl in $^5C_1$ ways and arrange them in $4!$ ways. Now we got three things: One is a group of $4$ girls, a group of $5$ boys and a lonely girl. Consider the group of $4$ girls as one object or item for a moment. You can arrange the $5$ boys in $5!$ ways. After arranging them you'll see that there will be $6$ spaces formed between those $5$ boys.
Hence if we chose $2$ places (one for the lonely girl and other for the group of girls) and arrange them between themselves we'll get $m$. Hence $m$ would be $5 \cdot 4! \cdot 5! \cdot$ $^6P_2$
