Arrangements in which only two of the three empty chairs are next to each other While studying, I got stuck on this problem: "Seven identical chairs in a row are to be seated by four students. How many arrangements are there such that the only two of the three empty chairs are next to each other?"
The answer key showed the answer as being 480, but I got 720. Could anyone explain how to get the correct answer of 480?
 A: Let's choose the empty chairs first:
Suppose they are at either end of the row (the empty chairs are 1 and 2, or 6 or 7... 2 possibilities). There are 4 ways of choosing the third empty chair, because the seat immediately next to the two empty chairs must be seated by a student.
Suppose the two empty chairs are somewhere in the middle of the row (the empty chairs are 2 and 3, or 3 and 4, or 4 and 5, or 5 and 6... 4 possibilities). Then the chair to the either side of the two empty chairs must be seated by students, and there are three chairs to choose from for the next empty chair.
So there are $2*4+4*3=20$ ways to choose the three empty chairs.
Now, for each of the 20 ways, there are 4 chairs that the students must be seated in. There are $4!=24$ ways to arrange 4 students in those 4 chairs.
By the multiplication principle, the total number of ways is $20*24=480.$
A: I think you should learn another method, which you may find simpler. 


*

*Club 2 empty chairs to be together as one big chair. So  empty chairs now form  2 objects.

*Seat the students in 4 chairs in 4! = 24 ways

*The 2 empty chairs (one normal, one big) can be placed in the 5 gaps between the students (including the ends) in $^5P_2$ = 20 ways

*Multiply to get ans = $24\times 20$ = 480  
