# Thread: tricky permutations problem

1. ## tricky permutations problem

This is very tricky.

8 people; A, B, C, D, E, F, G, and H sit around a table. How many ways can this be done if A must be directly opposite to B and C cannot sit next to D?

I had 912. it must be wrong

Thanks

2. Originally Posted by differentiate
8 people; A, B, C, D, E, F, G, and H sit around a table. How many ways can this be done if A must be directly opposite to B and C cannot sit next to D?
Seat A at the table. That fixes where B must sit. It also leaves six chairs empty.
There are four ways to seat C next to either A or B, leaving four ways to seat D.
Sitting C next to neither A nor B can be done in two ways leaving three ways to seat D.
So there are a total of twenty-two ways to seat C&D.
Now seat the rest.
So what is the answer?

3. Hello, differentiate!

Eight people: $\displaystyle \{ A, B, C, D, E, F, G, H\}$ sit around a table.
How many ways can this be done if $\displaystyle A$ must be directly opposite $\displaystyle B$
and $\displaystyle C$ cannot sit next to $\displaystyle D$?

I had 912. It must be wrong. .
How do you know?
I will assume it is a circular table.

$\displaystyle A$ can sit anywhere; it doesn't matter.
Then $\displaystyle B$'s seat is already determined.

So far, we have these 8 chairs with two of them occupied.
Code:
          A
/   \
X       X
|       |
X       X
|       |
X       X
\   /
B

Remove two of the chairs.
Place stools between the remaining six chairs.
Code:
          A
o   o
X       X
o       o
X       X
o   o
B

Seat $\displaystyle C$ in any of the 6 empty stools.
Seat $\displaystyle D$ is any of the 5 remaining stools.
. . There are: .$\displaystyle 6\cdot5 \,=\,30$ ways.

Then seat $\displaystyle E,F,G,H$ in the four chairs.
. . There are: .$\displaystyle 4! \,=\,24$ ways.

Therefore, there are: .$\displaystyle 30\cdot24 \:=\:720$ ways.

4. Originally Posted by differentiate
This is very tricky.

8 people; A, B, C, D, E, F, G, and H sit around a table. How many ways can this be done if A must be directly opposite to B and C cannot sit next to D?

I had 912. it must be wrong

Thanks
I'm getting a different answer from Soroban's, and since I don't understand his method with the stools, here is my approach.

Fix A and B and number the other chairs according to the following diagram:
Code:
            A
1      6
2            5
3     4
B
(That's supposed by be a circle, although of course it isn't. You have to use your imagination.)

If we disregard the restriction on C and D, there are 6! ways to place C-H.

Let's see if we can count the number of arrangements in which C does sit next to D. We start by placing C. If C is in seat 1, 3, 4, or 6, there is only one place for D to sit and then 4! ways to place E-H. If C is in seat 2 or 5 then there are 2 places for D to sit and then 4! ways to place E-H. So all together there are 4 x 4! + 2 x 2 x 4! = 8 x 4! ways to arrange C-H with C sitting next to D.

So the number of circular arrangements with A opposite B and C not sitting next to D is

6! - 8 x 4! = 528.

5. to be honest, I don't know what the answer is
I just want to know what you guys think.

6. Originally Posted by differentiate
to be honest, I don't know what the answer is
I just want to know what you guys think.
I understand your confusion.
My edited answer is correct. Both of the other replies have logical mistakes in them.
Look at the attached diagram. Seating A&B “fixes the table”.
If we seat C in any of 1,3,4,6 positions we can put D in any of four other positions.
However, if we seat C in either 2 or 5 then there are only three ways to seat D.
Therefore, once A&B are seated there are only 22 ways to seat C&D.
There are four left to be seated. That can be done in $\displaystyle 4!$ ways.

7. Originally Posted by Plato
I understand your confusion.
My edited answer is correct. Both of the other replies have logical mistakes in them.
Look at the attached diagram. Seating A&B “fixes the table”.
If we seat C in any of 1,3,4,6 positions we can put D in any of four other positions.
However, if we seat C in either 2 or 5 then there are only three ways to seat D.
Therefore, once A&B are seated there are only 22 ways to seat C&D.
There are four left to be seated. That can be done in $\displaystyle 4!$ ways.
Ahem!

Well, logical mistake or not, Plato and I seem to have arrived at the same number: 22 x 4! = 528.