Number of ways in which $5$ boys and $5$ girls can stand in a queue such that exactly $4$ girls can stand consecutively is There are $5$ boys and $5$ girls. Number of ways in which $5$ boys and $5$ girls can stand in a queue such that exactly $4$ girls can stand consecutively is:
My attempt:
I first chose $4$ girls out of $5$ needed to be arranged in queue. 
It can be done in $5C4$ ways. I bundle them separately.
Then I arrange them with remaining $5$ boys and a girl. 
So total ways is $5C4×4!×7!$
Now we have to subtract those cases in which all $5$ girls are together.
So ways are $6!×5!$
Required ans is $5C4×4!×7!-6!×5!$. 

Why this is wrong?

 A: There is double counting in $\binom54\times4!\times7!$ where you count the number of ways where at least $4$ girls are consecutive.
Suppose you have selected the girls $A,B,C,D$ to be consecutive.
Then we encounter the possibility: $$(A-B-C-D)-E-boys$$ where  $E$ is the other girl.
But if you have selected the girls $B,C,D,E$ we meet this possibility again in: $$A-(B-C-D-E)-boys$$

edit (solution)
First place the girls on a row.
Then by using stars and bars we find $2\binom62=15$ to place the boys in such a way that the girls are split up in a consecutive group of $4$ and a single girl. Here factor $2$ is there because the single girl can be on the left or on the right of the consecutive group.
That gives: $$2\binom625!5!=432000$$ possibilities in total
A: First, there are exactly 7 ways to choose positions for consecutive girls: (1,2,3,4), (2,3,4,5) ... (7,8,9,10). For each of these 7 options you can select 4 girls in 5 different ways (5 choose 4), but we do care about the order, so we need to multiply by 4!. For two "corner" options 1,2,3,4 and 7,8,9,10 remaining girl can be placed in 5 positions (excluding 5th position in the first variant and 6th in the second); for other options a remaining girl can only be placed in 4 positions (e.g., for 2,3,4,5 option a remaining girl cannot take 1st or 6th position). Number of ways to arrange remaining boys is 5!. Therefore we have:
2*5*4!*5*5! + 5*5*4!*4*5! = 5*5*5!*4!*6 = 432000. 
