Solved

Polya counting theory doubt

Posted on 2008-10-19
6
309 Views
Last Modified: 2012-08-14
saw a solution for the following problem in a book :
 
Problem :
Suppose there are 6 chairs around a circular table.
3 men and 3 women need to be seated around it.
Assuming that two sittings are equivalent when one
can find the other by means of a clockwise rotation
through 60k degrees 0<=K<=5,
how many different sittings are possible?
 
Solution :
There are 6 permutations possible : (set of vertices = {1,2,3,4,5,6})
P1 = (1) (2) (3) (4) (5) (6)
P2 = (1 2 3 4 5 6 )
P3 = (1 3 5 ) (2 4 6)
P4 = (1 4) (2 5) (3 6)
P5 = (1 5 3) (2 6 4)
P6 = (1  6 5 4 3 2)
 
So the cycle index is :
(1/6)*(x1^6 + x6 + x3^2 + x2^3 + x3^2 + x6)
 
Since number of men and women are equal, weight of both the colors is 1,
i.e. W(Men) = W(Women) = 1,
so putting x1 = 1+1, x2 = 1^2+1^2, x3 = 1^3 + 1^3 etc.
answer is
(1/6)*(2^6 + 2^3 + 2*2^2 + 2*2) = (1/6)*(64+8+2*4 + 2*2)
= (1/6)*(64+16+4) = (1/6)*84 = 14
--------------------------------------------------------------------------------------
 
Now, here the weights assigned for men and women are 1, since number of men = number of women = 3.
What if number of men = 4, number  of women = 2,
what would be the weights then?
What is the procedure to find out the weights in general?
0
Comment
Question by:dtivmk
  • 3
  • 3
6 Comments
 
LVL 84

Expert Comment

by:ozo
ID: 22755303
if number of men = 4, number  of women = 2, the possible permutations would change
0
 
LVL 1

Author Comment

by:dtivmk
ID: 22755321
Yes, they will.
So, how will the solution above change? that's the question...
Even in the above solution, premutations have been computed
without taking into account number of men/women...
They were accounted for only while calculating weights..
0
 
LVL 84

Accepted Solution

by:
ozo earned 500 total points
ID: 22755459

(1/6)*((x+y)^6 + (x^6+y^6) + (x^3+y^3)^2 + (x^2+y^2)^3 + (x^3+y^3)^2 + (x^6+y^6)
=
(
(y^6+6xy^5+15x^2y^4+20x^3y^3+15x^4y^2+6x^5y+x^6)
+
2(x^6+y^6)
 +
2(x^6+2x^3y^3+y^6)
 +
x^6+3x^4y^2+3x^2y^4+y^6
)
/6
=
(1y^6 + 1xy^5 + 3x^2y^4 + 4x^3y^3 + 3x^4y^2 + 1x^5y + 1x^6)
so
1 way to do it with 6 women
1 way to do it with 1 man 5 women
3 ways to do it with 2 men and 4 women
4 ways to do it with 3 men and 3 women
3 ways to do it with 4 men and 2 women
1 way to do it with 5 men 1 woman
1 way to do it with 6 men
total of 14




0
Free Tool: ZipGrep

ZipGrep is a utility that can list and search zip (.war, .ear, .jar, etc) archives for text patterns, without the need to extract the archive's contents.

One of a set of tools we're offering as a way to say thank you for being a part of the community.

 
LVL 1

Author Comment

by:dtivmk
ID: 22755524
Ok,
so here it's assumed that the all men are indistinguishable from each other.
what if they are not?
can polya's theory deal with that scenario?
0
 
LVL 84

Assisted Solution

by:ozo
ozo earned 500 total points
ID: 22755667
for 6 distinct people
(1/6)*((u+v+w+x+y+z)^6 + 2*(u^6+v^6+w^6+x^6+y^6+z^5) + 2*(u^3+v^3+w^3+x^3+y^3+z^3)^2 + (u^2+v^2+w^3+x^2+y^2+z^2)^3 )
where you have 1 of each person
so we are only intersted in the u*v*w*x*y*z term =
(1/6)*(... + 720*u*v*w*x*y*z + ... )
= 120 = 5! as expected
0
 
LVL 1

Author Comment

by:dtivmk
ID: 22765217
Just one question.
What is the concept of weights then?
Here weight of every color is 1.
What else could it possibly be?
0

Featured Post

Announcing the Most Valuable Experts of 2016

MVEs are more concerned with the satisfaction of those they help than with the considerable points they can earn. They are the types of people you feel privileged to call colleagues. Join us in honoring this amazing group of Experts.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Foreword (May 2015) This web page has appeared at Google.  It's definitely worth considering! https://www.google.com/about/careers/students/guide-to-technical-development.html How to Know You are Making a Difference at EE In August, 2013, one …
The greatest common divisor (gcd) of two positive integers is their largest common divisor. Let's consider two numbers 12 and 20. The divisors of 12 are 1, 2, 3, 4, 6, 12 The divisors of 20 are 1, 2, 4, 5, 10 20 The highest number among the c…
Although Jacob Bernoulli (1654-1705) has been credited as the creator of "Binomial Distribution Table", Gottfried Leibniz (1646-1716) did his dissertation on the subject in 1666; Leibniz you may recall is the co-inventor of "Calculus" and beat Isaac…
I've attached the XLSM Excel spreadsheet I used in the video and also text files containing the macros used.

791 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question