Number of seating arrangements such that no $2$ boys are together There are $5$ girls and $3$ boys and I need them to get seated in a row such that no 2 boys are together.
This is my attempt.
The total number of arrangements (without any condition) should be $8!$.
Now I find the arrangements in which two particular boys call them $A$ and $B$ are together. The number of ways that can be done is $7! \times 2!$.
Since I have $3$ boys, the number of choosing $2$ boys are $^3C_{2}$.
Hence the total number of ways in which  $2$ boys are always seated together are $7! \times 2!$ $\times$ $^3C_{2}$.
Hence the total number of ways in which arrangement can be done such that no $2$ boys are together are $8!$ - ($7! \times 2!$ $\times$ $^3C_{2}$)
I don't know why is this approach incorrect.
Please help.
 A: While $7! \cdot 2!$ counts the number of ways of boys $A$ and $B$ seated together, it does not restrict that $C$ is seated next to them on either side. In other words, it counts arrangements when all three of them are seated together.
Now here comes the problem -
The moment you multiply $7! \cdot 2! \ $ by $\displaystyle {3 \choose 2}$ to choose two boys from $A, B, C$ to be seated together, they together double count arrangements where all three of them are seated together.
You can subtract overcounting by subtracting $3! \times 6!$.
So the correct answer should be,
$ \displaystyle 8! - 3 \cdot 7! \cdot 2! + 6! \cdot 3! = 14400$
But there is a simpler way to go about it.
There are $5!$ ways to seat the girls and then boys can be seated in any of the $3$ out of $6$ spaces between girls (including two at the ends). That is $ \ \displaystyle 5! \cdot {6 \choose 3} \cdot 3! \ $ arrangements.
A: Your analysis is wrong (although a first step to a correct approach) because you are double-counting arrangements in which all three boys sit together.  For example, if $A, B, C$ are the boys and they sit together in that order, you're subtracting them once because $A$ and $B$ are sitting together, and a second time because $B$ and $C$ are sitting together.
If you apply the principle of inclusion and exclusion, you will see that you must add back all of those arrangements.  Thus, the correct answer is $8! - 7! \cdot 2! \cdot \binom 32 + 6! \cdot 3!$.
