There are 8 chairs around a round table. If A and B must sit together, then in how many ways A,B,C,D and E can sit around this table?
According to me : 4!.2! but I think I am forgetting something.
What makes this problem different is the eight chairs but only five people.
Seat A anywhere at the table. Now the tabled is ordered.
There are two places to seat B, on A's right or left.
There are then six places to seat C.
There are five places to seat D.
There are four places to seat E.
Because this is a "round table" and shifting every one to the left or right gives the same "ordering" we can set A anywhere at the table. Once we have done that, we can seat B on A's left or right- 2 choices. There are now 6 chairs left. We can seat C in any of those 6 chairs, then D in any of the 5 chairs left, then E in any of the 4 chairs left. There are 6(5)(4)= 120 ways to do that so 6(120)= 720 ways to seat these 4 people, with A and B sitting together. More formally, 6(5)(4)= 6!/3! so we could also write this as (2)(6!/3!).
Did you read reply #4?
Seat A anywhere at the table. Now the tabled is ordered.
There are two places to seat B, on A's right or left.
There are then six places to seat C.
There are five places to seat D.
There are four places to seat E.
$\displaystyle 2\cdot 6\cdot 5\cdot 4$