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
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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?
Ans :
(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
Also ;
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
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
=
(
(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
Also ;
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
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
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