Application of inclusion-exclusion principle 
There are 5 boys and 4 girls in my class. 
In how many ways can they be seated in a row of 9 chairs such that at
  least 3 girls are all next to each other?

I tried to solve using PIE as follows:
First count the number of ways to get 3 girls next to each other.
This is 24*6!*6 = 103680.
Then count the number of ways to get 4 girls next to each other.
This is 24*6! = 17280.
Each of the ways to get 4 girls is counted 4C3 in the 3 girls case, so you do:
103680 - (4C3)*(17280) = 34560.
However, this is incorrect and the answer should be 103680. Where did i go wrong?
 A: Hint: Use a technique called "grouping." In it (pun intended), you would group the three girls into a single object, resulting in 5 boys, 1 "super-group", and 1 girl. 
Treat the "super-group" as a single unique entity, and at the end when you have the number of ways that 5 boys, 1 "super-group", and 1 girl can arrange themselves (7!), multiply it by 4P3 (because we don't know which 4 girls will be in the group). Almost looks good. Where did we overcount?
I will let you determine that. As another hint, subtract the current overcounted number from the solution. See if you get a number that looks familiar.
A: Here is an alternate approach that avoids overcounting:
First seat the 5 boys in a row, which can be done in $5!$ ways.
$\textbf{A)}$ If all 4 girls sit together, then there are 6 choices for the gap (between the boys) in which they sit, and $4!$ ways to arrange the girls in a row; so this gives $5!\cdot6\cdot4!=17,280$ possibilities.
$\textbf{B)}$ If only 3 girls sit together, then there are 6 choices for the gap for the 3 girls, $\binom{4}{3}$ ways to choose the 3 girls, $3!$ ways to arrange them in a row, and 5 choices for the gap for the remaining girl; so this gives $5!\cdot6\cdot\binom{4}{3}\cdot3!\cdot5=86,400$ possibilities.
Therefore there are $17,280+86,400=103,680$ possibilities altogether.
