Quick Permutations Combinations Question (1 Viewer)

[Darrow]

Ok, You have 8 people (4 couples) who are arranged around a circle
Now, the couples have to be opposite from each other
So, how many ways can they be arranged?

2 Thoughts
Its going to be 3! x (2!)^ (Either 3 or 4)

My question is do we assume the table to be distinct, that is, have a top end and bottom end, that way if every couple switched sides it will be different (and thus 2!^4)

or

The table is considered relative to the couples and if everyone switched it will be the exact same as if none of them switched (thus 2!^3)

Any thoughts?

Edit: Sorry, I did mean it to be 4 couples 8 people

 

[bored of sc]

The table is considered relative to the couples and if everyone switched it will be the exact same as if none of them switched (thus 2!^3)

True.

I would do the following:

Treat each couple as one which can be arranged 2! ways.

Thus, you have 8 x 2! = 16 ways.

That's wrong isn't it?

 

[lyounamu]

2 x 4 x 6! = 8 x 6!

 

[lyounamu]

True.

I would do the following:

Treat each couple as one which can be arranged 2! ways.

Thus, you have 8 x 2! = 16 ways.

That's wrong isn't it?

You were actually on the right track but why did you times 8 by 2!?

It should really be 4 x 2! because if you have 8 x 2! you are actually doubling it. Draw a circle and see, you will see that there are 8. But other 6 people are arranged in 6! ways, so 6! x 8 = lasjasldj (I don't have a calculator here)

 

[gh0stface]

8! x 2! from the way i interpret ur question?



theres 8 couples, so 16 people rite? each couple is opposite to another so theres 2 people facing another 2 people rite?

 

[lolokay]

I would say 3! * 2!³
(since it's a circle, there's no top and bottom end)

 

[lyounamu]

Ah was it 8 couples? I thought it was 8 PEOPLE. Damn...my eye sight is getting worse.

 

[lolokay]

haha I read it as 8 people aswell ^

wouldn't the answer then be 7!*2⁷ ?

 

[lyounamu]

haha I read it as 8 people aswell ^

wouldn't the answer then be 7!*2⁷ ?

Yeah. I concur.

 

[whoisurdaddy]

it is 16 right? Since the other couples dont affect the number of arrangements.

 

[Darrow]

Sorry, I did mean it to be 8 people 4 couples
My thoughts were:
Treat the 8 people as 4 groups (just like if they had to sit next to each other, because as one person sits, their partner has a predetermined position)
But being a circle someone has to be the beginning (and their partner) so it becomes

3!

But each couple can switch sides, ie, Wife can be at the top Husband at the bottom, or Husband at the top, wife at the bottom

So, 3! x 2! x 2! x 2! x2! = 96
But the answer sheet (which could be wrong) says its only
3! x 2! x2! x2! = 48

Now, who can justify the latter answer?
I have a feeling it has something to do with the circle, if all couples switch sides, then you have the exact same arrangement, so its discounted by using the first couple as a distinct reference

 

[lolokay]

"3! x 2! x2! x2! = 12"
um, doesn't it = 48?

you do have the exact same arrangement if all couples switch sides - your 'someone [at] the beginning' isn't supposed to move, or else you would have to allow for all people moving 1 to the right etc.etc.

 

[tommykins]

回复: Re: Quick Permutations Combinations Question

Stick a couple in a spot.
There are 2! ways to arrange this couple. Since there are 4 couples, we have 2!^4 ways of arrangement.

4 groups in a circle, there are 3! ways to organise these groups.

Total = 3!x2!^4, but if we switch the couples it's still the same couple sitting opposite to each other, hence divide by 2 (which is 2!)
Total arrangements = 3!x2!^3

 

[Darrow]

Re: 回复: Re: Quick Permutations Combinations Question

Tommykins,
The problem with that line of thought is do we accept the circle as relative?
If the first couple are allowed to switch positions, we still have our reference as Couple 1
If all couples swap, shouldn't it be different from if none of them swap?

Your reasoning would assume that we can rotate the table to make it the same

 

[adosh]

can someone please help me with permutations and combinations....when do we use nCr and nPr ,,how can we know which to use for a question??? thanks