In how many ways can the 10 kids in my class be seated in a circle if John and Sam insist on being seated diametrically opposite each other and if Stacy and Roger also insist on being diametrically opposite each other?

Guest Mar 14, 2015

#4**+10 **

Thanks Alan and CPhill

Okay, after consideration I also agree with CPhill and Alan.

It should be noted that we are all assuming that the position in relation to the room is irrelevent.

This is my version of the logic. (I may just be repeating what Alan or Chris as already said)

No disrespect intended. :)

1) Like Alan did I am going to sit John and Sam first. It doesnt matter where so long as they are opposite each other.

2) Now I am going to forget about Stacy and Roger - **they are late to class**!

3) there are 6 people and 6 positions that they must occupy. That is 6!

4) **Stacy just turned up.** We better squeeze her in some place. There are 8 places where she can squeeze in.

5) Now Roger saunters into class. He has to sit opposite Stacey so there is only one place he can sit.

HENCE: The number of permutations is 6!*8*1 = **6!*8**

Melody
Mar 15, 2015

#2**+10 **

Let John define position 1, so Sam defines position 6.

1. Put Stacey in position 2, then Roger is in position 7. There are 6! ways of arranging the others.

2. Switch Stacey and Roger. There are another 6! ways of arranging the others.

3. Put Stacey in position 3, then Roger is in position 8. There are 6! ways of arranging the others.

4. Switch Stacey and Roger. There are 6! ways of arranging the others.

5. Put Stacey in position 4, Roger in position 9. There are 6! ways of arranging the others.

6. Switch Stacey and Roger. There are 6! ways of arranging the others.

7. Put Stacey in position 5, Roger in position 10. There are 6! ways of arranging the others.

8. Switch Stacey and Roger. There are 6! ways of arranging the others.

Hence, in total there are 8*6! arrangements.

$${\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{6}}{!} = {\mathtt{5\,760}}$$

.

Alan
Mar 14, 2015

#3**+15 **

I'll give this one a shot.......

"Anchor" John and Sam in any two opposite positions. Now, Stacy and Roger can occupy any of the other 4 diametrically opposite positions in 2 ways each, and for each of these arragements the other children can be further arranged in 6! ways.

So we have 4 x 2 x 6! = 5760 ways

Hey...Alan and I agree......whether that makes us correct - or not -might be another story..!!!

CPhill
Mar 15, 2015

#4**+10 **

Best Answer

Thanks Alan and CPhill

Okay, after consideration I also agree with CPhill and Alan.

It should be noted that we are all assuming that the position in relation to the room is irrelevent.

This is my version of the logic. (I may just be repeating what Alan or Chris as already said)

No disrespect intended. :)

1) Like Alan did I am going to sit John and Sam first. It doesnt matter where so long as they are opposite each other.

2) Now I am going to forget about Stacy and Roger - **they are late to class**!

3) there are 6 people and 6 positions that they must occupy. That is 6!

4) **Stacy just turned up.** We better squeeze her in some place. There are 8 places where she can squeeze in.

5) Now Roger saunters into class. He has to sit opposite Stacey so there is only one place he can sit.

HENCE: The number of permutations is 6!*8*1 = **6!*8**

Melody
Mar 15, 2015

#5**+5 **

OH NO!!!.....It's the dreaded "Octagon" again.....!!!!

Hey....where is that border around it????

Did you steal it, Melody???

CPhill
Mar 15, 2015

#7**0 **

I am so sorry, but I put this answer in and it said it was wrong. Sorry again because I still don't know so I won't be able to explain it.

Guest Mar 15, 2015

#8**0 **

wait, not sorry. I think there was another question similar to this one, but got deleted. They were very similar it's just that one part was different.

Here it is if someone would like to help me please:

In how many ways can the 10 kids in my class be seated in a circle if John and Sam insist on being seated diametrically opposite each other? (As usual, two seatings which are rotations of each other are considered the same.)

Guest Mar 15, 2015

#9**+5 **

Well you have been given lots of information here.

What do you think the answer might be? How will you try to do it?

It is very similar to this question - just a little easier.

We are not here just to provide answers so that you can plug the answer in and get it right.

If you want more input on this question you should work with us :)

Melody
Mar 16, 2015