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# In how many ways can 8 people be seated around a square table with 2 people on a side? (Two configurations are considered equivalent if one

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In how many ways can 8 people be seated around a square table with 2 people on a side? (Two configurations are considered equivalent if one is a rotation of another.)

Guest Apr 1, 2015

#20
+889
+10

3*5*7*...*(2k-1)

= 2*3*4*5*6*7*...*(2k-2)*(2k-1) / (2*4*6*...*(2k-2))

= (2k-1)! / (2^(k-1)*1*2*3*...*(k-1))

= (2k-1)! / (2^(k-1)*(k-1)!)

Bertie  Apr 5, 2015
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#1
+81142
+5

One person can be 'anchored" in one of two seats on one side.......the other seven people can be arranged in 7! ways.....so....

2 x 7!  =

2 x 5040 = 10,080 ways

CPhill  Apr 1, 2015
#2
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This post  is all nonsense -

I am just leaving it here so that Alan's and Bertie's comments make sense.

I think you misread the question Chris,

And I think it might be  (table rotations are considered the same)

8C2*6C2*4C2*2C2*(4-1)!

$${\left({\frac{{\mathtt{8}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{6}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{4}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{3}}{!} = {\mathtt{15\,120}}$$

This seems much to big

Chris's answer - where no one was pair up - was    $${\mathtt{7}}{!} = {\mathtt{5\,040}}$$

And i thought my answer should be smaller than this.

Melody  Apr 1, 2015
#3
+81142
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Here's the justification for my answer....

Consider a "birdseye" view of the table....

One person either occupies the left seat on one side of the table or the right seat on that side.

And for each of these arrangements, the other 7 seats can be filled in all possible ways = 7! ways.

So, 2 x 7!   = 10,080 ways

"Rotations" of the table don't matter......one rotation would look like any other for a particular arrangement.

CPhill  Apr 1, 2015
#4
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I like the working of my answer but I don't understand why  the number would be so much bigger (or bigger at all) than if everyone could just sit where they wanted.

Both your answer and mine have a bigger value than if everyone was individuals and to me this seems a little crazy.

There has to be doubling up happening.  !!

Melody  Apr 1, 2015
#5
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Hi Chris and Melody, hope you don't mind me joining the conversation.

I'm with Chris on this one, I think that the 10080 is correct, (I've gone through the calculation in a different way and arrived at the same answer).

So, here's a question for Melody. Forget about rotations and seat positions for the moment, suppose that I have eight people and that I'm going to split them into four groups of two. In how many ways can this be done ?

Let's see If we can get agreement on that to begin with.

Bertie  Apr 2, 2015
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Hi Bertie, I was hoping that you would join in. thank you.

So, here's a question for Melody. Forget about rotations and seat positions for the moment, suppose that I have eight people and that I'm going to split them into four groups of two. In how many ways can this be done ?

Well I thought that it was

8C2*6C2*4C2*2C2 but this is obviously not correct - I guess i am double counting again :/

Melody  Apr 2, 2015
#7
+26412
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It might help to think of a simpler situation first.  Consider four people sitting two each side of a two-sided table(!).

The distinct possibilities, excluding rotations, are easily visualised:

.

Alan  Apr 2, 2015
#8
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Okay thanks Alan, I think I have digested that.  And I can see why Chris's answer makes sense.

BUT

I still would like to understand why my logic  is wrong :)

I have always thought with probability that it is often fairly easy to understand the correct answer but much harder to understand why an incorrect answer is wrong.

Still, with all the prob questions we have been getting I think my ability is improving :)

Melody  Apr 2, 2015
#9
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Worse than double counting, multiple counting.

To illustrate why this method is wrong, consider an even simpler situation. Suppose that you have just four people and you want to split them into pairs. How many ways ?

If you employ the method used for the eight person problem, I guess that you would say that the number would be 4C2*2C2 = 6.

However, if you call the people A,B,C,D, you can see that there are only three possible pairings (AB and CD), (AC and BD) and (AD and BC). The basic reason why the first answer is wrong is that each grouping has been counted twice. (AB and CD) and (CD and AB), for example, are being thought of as being two different pairings, meaning that that particular pairing has been counted twice, if that makes sense. The 4C2 = 6 contains AB, AC, AD, BC, BD and CD, but both AB and CD produce the same split into two pairs.

Try the eight into four groups of two again.

Bertie  Apr 2, 2015
#11
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Okay I am just starting to get somewhere.

I am looking at how many ways even numbers of people can be paired off

This post is NOT directly related to the question above.

4 people   ABC and D

AB CD

AC BD

that is 4 people can be paired off in 3 ways

Now, before I wanted to do it as 4C2 which is 6

This would give the 6 above but each time you get one pair you automatically get the another pair so it can be seen that this will be too big by a factor of 2   so it is   4C2/2 = 6/2=3

---------------------------------------------------------

6 People

Now I am going to look at 6 people  A B C D E and  F

AB    there  are four more so they can be paired up 3 ways

AC    3 ways

AE     3 ways

AF     3 ways

Total = 3*5 =15

So 6 people can be paired off  15 ways.   This is  (6-1)(4C2)/2  = 15

----------------------------------------------------

8 People

Now I am going to look at 8 people  A B C D E F G and H

AB   then 6 more =  15 ways

AC   then 6 more =  15 ways

AD   then 6 more =  15 ways

AE   then 6 more =  15 ways

AF   then 6 more =  15 ways

AG   then 6 more =  15 ways

AH  then 6 more =  15 ways

That is a total of  7*15=105    (8-1)(6-1)4C2/2 = 7*5*3

that is 8 people can be paired off in 105 ways

-------------------------------------------------

So   2K people can be paired off in   3*5*7*......*(2K-1)   ways

------------------------------------------------------

So   4 people are paired (and the pairs are then considered as 1)

they are then put clockwise into a cirlce (or anticlockwise)  rotations are considered the same.

The number of ways this could be done is  4*1=4 ways

6 people - it could  be done   5*3*2! = 30 ways

8 people - it could be done  7*5*3*3! = 1050 ways

2K people - it could be done  (2k-1)(2k-3)....5*3*(k-1)!

-----------------------------------------------------------

Melody  Apr 4, 2015
#12
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Look Bertie, Alan and Chris, I did it my way and I got the same answer as Bertie and Chris!      WOW!!

I know my way is no where near as sensible as Chris's way but I learned heaps doing it this way

4 people

There are 3 ways to pair up 4 people, if the order of the pairs counts then this would be 3*2*2 = 12 ways

For instance   AB CD could also be AB DC, BA CD, or BA CD

They can then be put in a circle 12*(2-1) = 12 ways

So 4 people can be paired off and THEN placed in a circle 12 ways

----------------------------

6 people

There are 3*5 ways to pair up 6 people, if the order of the pairs counts then this would be 15*2^3 = 120 ways

Now they can be placed in a circle   120*2! = 240

So 6 people can be paired off and THEN placed in a circle 240 ways

--------------------------------

8 people

There are 3*5*7=105 ways to pair up 8 people, if the order of the pairs counts then this would be 105*2^4 = 1680 ways

Now they can be placed in a circle   1680*3! = 240

So 8 people can be paired off and THEN placed in a circle 10080 ways         @@  BINGO  @@

Melody  Apr 4, 2015
#13
+91510
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Now I have a new question.

2K people can be paired off in   3*5*7*......*(2K-1)   ways

How can I write this as a formula?

Melody  Apr 4, 2015
#14
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+10

Possibly as:

$$\prod_{n=2}^K(2n-1)$$

or as:

$$\frac{2^{K+1}\Gamma(K+\frac{1}{2})}{2\sqrt{\pi}}$$

where Γ() is the Gamma function

$$\Gamma (t)=\int_0^{\infty}x^{t-1}e^{-x}dx$$

.

Alan  Apr 4, 2015
#15
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+5

Thanks,

I don't know that symbol Alan, I thought that there must be one to do that job.

What is the symbol called?   :/ [I am talking about the first formula]

Gamma function ?  mmm.....

Melody  Apr 4, 2015
#16
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Π is the Greek upper case letter, pi, and stands for the product (analogous to Σ for the sum).

Alan  Apr 4, 2015
#17
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Ok thanks Alan.

That gamma function.  I know I have seen it before but could you refresh my memory please?

Melody  Apr 4, 2015
#18
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The gamma function is a sort of continuous version of the factorial function.

When n is an integer then Γ(n) = (n-1)!

But Γ(n) is also defined for non-integer values of n.

.

Alan  Apr 4, 2015
#19
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Another answer for me to chew on - thanks Alan :)

Melody  Apr 4, 2015
#20
+889
+10

3*5*7*...*(2k-1)

= 2*3*4*5*6*7*...*(2k-2)*(2k-1) / (2*4*6*...*(2k-2))

= (2k-1)! / (2^(k-1)*1*2*3*...*(k-1))

= (2k-1)! / (2^(k-1)*(k-1)!)

Bertie  Apr 5, 2015
#21
+91510
+5

Thanks Bertie,

You know I was really proud of myself getting that far on my own.

I chewed on it for ages!

I have repeated your work here because I was having problems with it but by the time I had written it all out in LaTex I finally understood,  but I decided just to leave the LaTex here.

-------------------------------------

So the number of ways that 2k people can be paired off is:

$$3*5*7*9*.....*(2k-1)\\\\ =3*5*7*9*.....*(2k-3)*(2k-1)\\\\ =\frac{1*2*3*4*.....*(2k-3)(2k-2)(2k-1)}{2*4*6*8*...(2k-2)}\\\\ =\frac{1*2*3*4*.....*(2k-3)(2k-2)(2k-1)}{2^{k-1}[1*2*3*4*...(k-1)]}\\\\ =\frac{(2k-1)!}{2^{k-1}(k-1)!}$$

THIS IS GREAT!

Can this be extended to seperating into other groups Like 3N separated into groups of 3?

A project for the future perhaps.

Melody  Apr 5, 2015
#22
+91510
+5

Now Alan, I want to look at yours :)

This first one is very obvious, thank you

$$\prod_{n=2}^K(2n-1)$$

or as:

Now this I don't  get.

$$\frac{2^{K+1}\Gamma(K+\frac{1}{2})}{2\sqrt{\pi}} =\frac{2^k(K-\frac{1}{2})!}{\sqrt{\pi}}\;\;?$$
Where did $$\sqrt{\pi}$$      appear from?

where Γ() is the Gamma function

$$\Gamma (t)=\int_0^{\infty}x^{t-1}e^{-x}dx$$

It doesn't really matter, I  think this is all be above me.

Melody  Apr 5, 2015
#22
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Hi All,

I was intrigued by the answer Melody gave and would like to comment on it. See the approach of diving people into groups of same sizes is a pretty standard one. For example if you want ot divide a group of 8 people into pairs (i.e. all equal sizes) the standard approach shall be 8!/(((2!)^4)*4!), i.e divide 8! by 2!*2!*2!*2! and multiply the whole with 1/4! (since there are 4 groups of equal sizes) and it returns the same answer as 105. This can be used for anything. (3N as well). Say N people need to be distributed in 2 groups of 2 and 1 group of 3, then it is N!/2!*2!*3!*2! (i.e. 2! -> since 2 items belon to one group, 2! -> since 2 items belon to one group,3! -> since 2 items belon to one group, 2! -> since 2 groups are identical in size.)

Now that concept out of the way, the thing that intrigued me is its application in geometrical figures, like the one discussed. I think it has happened because in such a scenario we fix one group namely AB and then form 3 paired groups, and since it is a square all the 8 arrangements will be identical, hence giving us the formula (2k-1)!/2^k-1*3!

i.e. (2k)!/(2k*((2!)^3)*3!)

I think that is the case. But definitely for sure such a formula wouldnt work out in other geometrical figures or if the numbers werent 8 for example, i.e we have 6 persons and there are 8 places.

Guest Nov 25, 2015
#23
+5

After all these discuession, i just wanna post the correct answer just to make it clear.

There are 8! ways to place the people around the table, but this counts each valid arrangement 4 times (if you move each person 2, 4, or 6 places clockwise you get the same arrangement). The answer is 8!/4 = 10080.

Guest Dec 28, 2015
#24
+91510
0

In how many ways can 8 people be seated around a square table with 2 people on a side? (Two configurations are considered equivalent if one is a rotation of another.)

Since this question has been dredged back up I decided to pay with it again (just to check that i can still do it ).......

It reall depends on what the questioner means.

If you are treating the square table like a round table then there are (8-1)! = 5040

BUT If you rotate just one place then the pairs are different so this number is doubled

5040*2= 10080

Our guest has just given you exactly the same solution, thanks for making it clear guest.

Our guests logic was a little different from mine but it is totalyy correct (both logic and answer)

------------------------------

If you only care about who is sitting in pairs and where the pairs are in relation to the otheraround the table (who is one the right or left does not matter) then the problem becomes different.

The answer would be who many pairs can be formed x how many ways can they be seated.

How many ways can the 4 pairs be seated = 3! =6

How many ways can they be fromed.

If there were only 2 people there would be 1 way.

If there were 4 people ABCD then AB, AC,AD  who A is with defines BOTH pairs so ther is 3 ways.

If people E and G join the mix  then if they are together then there are 3 ways.

Bur E could be with G or A or B or C or D   So that is 5 x3 ways.

By extention, if there are 8 people there are 7*5*3 = 105 ways they can be paired off.

I remember someone told me the notation is 7!!

So the number of ways pairs can be chosen and the pairs seated at the table (If AB are a pair that is the same as BA)  is 105*6 = 630 ways.

If where in a pair, personApersonB is different from personBpersonA the number would be much bigger. I think then the number would be back to 10,080 again as I think it would be the same as the original problem ://

Melody  Dec 28, 2015
edited by Melody  Dec 28, 2015

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