Cosplay Lineup!

If there's 5 guys, and there are 4 spaces required between them, then you would need 5 guys + 4 random dudes for the arrangement to work. Considering there is a total of only 8 guys, and 9 is needed, then this is impossible to achieve. Thus the answer is 0.

Had there been 9 people, I believe the answer would be 5!*4!=2880. (Since there's 5! ways to place the 5 guys with the same costume so there's a space in between each of them, and 4! ways to place the other dudes to make the said spaces.)

Instead of the same-character-playing people, we choose the other three. They can be placed in 3! ways. Now, there are four spaces left between them. We are supposed to place 5 people in 4 places, so by PHP, at least two of them will be in the same space,i.e, they will be next to each other. So the answer is 0.

Imagine the eight cosplayers sitting in a row of eight chairs (one person per seat). Consider two seats to be a "box." There are four boxes, but five people in the same costume. Hence, by Dirichlet, one box must contain two cosplayers. That is, two cosplayers in the same costume must sit next to each other. Hence, there are zero such arrangements.