Five math coaches A, B, C, D, E are seated in a row at a banquet. How many possible seatings are there, if the coaches A and D cannot be seated next to each other.
Note that the total possible arrangements are 5! = 120
Let's look at the arrangements where AD do sit together
AD - - -
-AD - -
- - AD -
- - - AD
So they could occupy 4 different positions....and for each of these they could be seated in 2 ways (AD or DA)
And for each of these.....the other 3 can be seated in 3! ways
So.....the number of seatingss where they don't sit together =
120 - (4)(2)(3!) =
120 - 8*6 =
120 - 48 =
72