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.

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- Jan 25th 2013, 03:12 AMkastamonuRound Table
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. - Jan 25th 2013, 05:07 AMibduttRe: Round Table
n persons can be made to sit on a round table is given by (n-1)! and two persons can be made to sit together in 2! ways. I am sure now you can reason out.

- Jan 25th 2013, 05:25 AMkastamonuRe: Round Table
Yes but there are 8 chairs. AB must sit together. This is 4!. They can be arranged beyween themselves. This is 2!.But something is missing.

- Jan 25th 2013, 05:58 AMPlatoRe: Round Table
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. - Jan 25th 2013, 07:07 AMHallsofIvyRe: Round Table
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!).

- Jan 25th 2013, 09:09 AMkastamonuRe: Round Table
But why did you multiply 120 with 6? How did you get 6?

- Jan 25th 2013, 09:19 AMPlatoRe: Round Table
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.

$\displaystyle 2\cdot 6\cdot 5\cdot 4$ - Jan 25th 2013, 10:46 PMkastamonuRe: Round Table
Many Thanks Plato. But why did HalsofIvy multyiply 120 by 6. This is what I couldn't make out.

- Jan 26th 2013, 04:24 AMPlatoRe: Round Table
- Jan 26th 2013, 09:52 AMkastamonuRe: Round Table
Many Thanks.