There are $9$ students of which $5$ are boys and $4$ are girls and we have to find the number of arrangements of all the $9$ students in such a way that all the $4$ girls do not come together.
I came this far:I know that there are $6$ places for each girl to sit on. But according to the question the can be three types of cases:
- when every girl sits alone
- when two girls sit together
- when $3$ girls sit together
I am assuming that the boys and girls are distinguishable.
The answer is very simple : in any arrangement of boys and girls, either : all four girls come together, or ; all the four girls don't come together.
Therefore, since you are trying to figure out the number of arrangements where the girls do not sit together, you must simply do this : find the total number of arrangements, and subtract from this the number of arrangements in which the girls sit together.
The total number of arrangements is $9!$, since you are arranging nine objects in total.
The number of arrangements in which all the girls sit together can be calculated like this : first, pick where the leftmost girl sits : she must sit in positions $1 \to 6$ if the positions are numbered $1,...,9$. Now, the positions of all the girls are fixed, and they can be permuted in these positions in $4!$ ways. Similarly, the positions of the other boys is fixed, and they can be permuted in $5!$ ways. Hence, the answer is $6 \times 4! \times 5! = 4! \times 6!$.
Therefore, the answer is $9! - 4!6! = 345600$. As you can see, this is a lot more than the $36$ you have written.
If the boys and girls are indistinguishable, then the total number of arrangements is $\binom 94$, since we are picking $4$ positions out of $9$ which the girls will occupy, and the number of disallowed positions will just be $6$ of them, since the girls and boys can't be permuted amongst themselves now. So the answer is $\binom 94 - 6 = 120$.