There are 7 candidates for the 4 consultant slots. If 2 out of 7 consultants refuse to be on the same team, how many different teams are possible?

Ill solve this using the slot method

Possible conditions with the constraints

7*6*5*4 = 35
4*3*2*1
When constraints are in place-- Lets assume that the 2 consultants are already selected,so what we need to do is to find the possible combinations for the other 2 consultants

5*4= 10
2*1

So 35-10=25 possible combinations where the 2candidates are not on the same team.


My concern is why I cant use the same method in this question below

Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many ways different arrangements can the 6 people sit?

So now we have 6 candidates Jan,Marcia,Greg,Peter,Bobby,Cindy

To find arrangements in which Marcia and Jan don't sit next to each other,I first find the number of arrangements in which they sit next to each other so the I can subtract those later from the total possible arrangements.

Lets assume J and M are already seated next to eachother,so for the remaining 4 spots we find the possible arrangements

4*3*2*1=24

Since order matters in which M and J sit next to each other opposed to J and M we found out earlier we multiply by 2 thus making it 48 possible arrangements.Therefore mt answer is 6!-48,but the answer is 6!-240=480 .

Why cant I use this method here??I know what the book says,but I want to know why Im wrong,I find it hard to remember too many things in P&C.