In how many ways can 8 people be seated in a row of chairs if three of the people, John, Wilma and Paul, refuse to sit in three consecutive seats?
I think it's easiest to count the ways that they can consecutively sit together and subtract this from the total number of seating arrangements.
There are 6 groups of 3 consecutive seats. Within each group there are 3!=6 ways the three fussy people can be arranged.
Then the remaining 5 people can be arranged 5!=120 ways in the remaining seats.
So there are 6*6*120 = 4320 arrangements where they DO sit consecutively
There are 8! = 40320 total arrangements.
So there are 40320 - 4320 = 36000 arrangements where the 3 do not sit consecutively.