Suppose there are 8 boys and 4 girls. Suppose they are lining up and entering the room in order e.g (b,g,b,b,g,b,g,b,b,g,b,b). The only rule is that the number of girls in the room cannot be greater than the number of boys. How many different combination of the entering order are there?
Editted (Attempts): First i try to consider $(x,y)$ represent $x$ girls and $y$ boys. Then I convert the problem into how many ways to travel from $(0,0)$ to $(4,8)$ with moving up or right only and not going below the line $y=x$. then i am not sure what are the right way to calculate
Let us not work with $8$ boys but with $9$ boys and let us demand that at any stage there must be more boys than girls in the room.
That gives evidently the same number of possibilities.
There are $\binom{13}4$ possibilities of lining up if the demand is neglected.
According to Bertrand's ballot theorem the number of possibilities in which at any stage there are more boys than girls in the room is: $$\frac{9-4}{9+4}\binom{13}4=275$$