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# In how many ways can five married couples be seated around a circle so that spouses sit together?

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In how many ways can five married couples be seated around a circle so that spouses sit together?

Guest Jul 22, 2014

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Don’t be so amazed, Melody. That happens every once in a while. LOL   Ditto!

...the M1 and F1 should still switch—shouldn’t they?   M1 must always stay in the same place, otherwise the result is the equivalent of a rotation of one of the other arrangements.

One thing about hyper-geometric distributions, they hyper-stretch one's brain.  Hear hear!  That's why I like to draw pictures where possible!

Alan  Jul 22, 2014
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#1
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Probability puzzles are never my forte. Oh well , better luck next time.

NO IT IS NOT CORRECT.  THANK YOU ANONYMOUS FOR THE CORRECTION

It seems the jury is out.  Who knows what is correct.

Since Alan is backing me I've gone back to thinking it may be alright.

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5 cojoined people can be seated at a table in 4! ways.

The first couple can sit anywhere

There are 4 choices for the next 2 seats

There are 3 choices for the next 2 seats etc

5 cojoined people can be sited at a table in 4! ways.

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now each of the five couples can have the wife on the left or on the right that is 2 ways.

So I think that the answer is 4! * 2^5= 24*32

$${\mathtt{4}}{!}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{5}}} = {\mathtt{768}}$$

Melody  Jul 22, 2014
#2
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It is probably

$$4!*((4*2)-1)=168$$

Some of these permutations may not be unique.

In any event, if it is a swingers party, they won’t stay that way!?

The exclamation mark with the question mark is the unknown permutation arrangement of how they will end up.

Guest Jul 22, 2014
#3
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Sorry made an error it is

$$4!*((5*2)-1)=216$$

Still not sure if all of these permutations are unique.They probably are.

Guest Jul 22, 2014
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Not that I want to overcomplicate things here, but, if a wife sits to the left of her husband does that count as a different arrangement from sitting to the right of him or not?

Ok.  Didn't read Melody's reply carefully enough!  Thanks for pointing this out Anonymous.

Alan  Jul 22, 2014
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The equations above your posting defines that in the affirmative.

If not, then the solution is 4! Where (n-1)! Is the basic circular permutation formula and the couples are (n).

(The !? part remains the same, though).

Guest Jul 22, 2014
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Thanks anonymous and hello Alan,

Yes I can instantly see the logic behind the 2*5 but I will have to think about the -1

Who  am I trying to kid - I do not have a clue

We should hire ourselves out to make plausable multilple choice answers for these types of questions.

(Sorry that was a bit offensive - maybe just I should hire myself out.  Maybe Chris would like to help me?)

I wonder if anyone would pay us for that?

Why is it 2*5-1?     I don't see any intrinsic logic behind the -1

Melody  Jul 22, 2014
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If rotations are not allowed, but reflections are, then I think Melody's formula is correct.  See the actual arrangements below for two couples and for three couples:

Alan  Jul 22, 2014
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WOW

Alan are you saying i was right!   That's amazing!

Melody  Jul 22, 2014
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Don’t be so amazed, Melody. That happen every once in a while. LOL

Without the minus 1, the first set would be counted twice. Consider after each “main” permutation the couples are switched. This is a binary switch, and is demonstrable by using 5 coins. Start with all heads up then flip each in sequence on the tenth sequence you have returned to the beginning –all heads up. I’m still not sure if all the combined permutations are unique.

Considering Alan’s graphic I need to study this more.

The original permutation formula 4!*((5*2)-1)=216 should allow for all defined permutations, including reflections, where the only restriction is couples sit next to each other.

I need to study your graphic for awhile. One note is the M1 and F1 do not switch, where they would using the above formula. Even with the restriction of reflection only, the M1 and F1 should still switch—shouldn’t they?

This is curious. . . . We know the (n-1)! for the couples placements is correct, so question is the counting of the “sub- permutation” where the couples switch places. I may run computer simulations to examine the uniqueness of the outcomes. The formula I offered may only work in certain situations.

One thing about hyper-geometric distributions, they hyper-stretch one's brain.

Guest Jul 22, 2014
#10
+26403
+5

Don’t be so amazed, Melody. That happens every once in a while. LOL   Ditto!

...the M1 and F1 should still switch—shouldn’t they?   M1 must always stay in the same place, otherwise the result is the equivalent of a rotation of one of the other arrangements.

One thing about hyper-geometric distributions, they hyper-stretch one's brain.  Hear hear!  That's why I like to draw pictures where possible!

Alan  Jul 22, 2014
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+91469
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My thanks to both of you.

Melody  Jul 22, 2014

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