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.

Guest Dec 20, 2019

#1**+1 **

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

CPhill Dec 21, 2019