Re: Bad understanding of GF


dxdydz wrote:
angela wrote:
Hi;

I am using a generating function to model a small partition problem like this:

From the set {1,2,3} with replacement how many ways can I sum to six with 3 choices. I use (x+x^2+x^3)^3 as the ordinary GF. The coefficient of x^6 is 7 so the answer is 7.

My problem is there are only 2 ways (1,2,3) and (2,2,2). The only way to get 7 is to treat (1,2,3) as 6 permutations and (2,2,2) as 1 permutation. I thought Ordinary GF were only Combinations and Exponential GF were for permutations. So why isnt the answer 2.

Gratefully,
Angela

I am not quite sure but the answer must be not to take the commutative
property of the multiplication as granted. Hence x*x^2*x^3 wont be the
same as x*x^3*x^2 or x^2*x*x^3 or x^2*x^3*x e.t.c. In the result keep
only one term from the terms that would be equal if the multiplication
commutative property would hold, then do the addition.
So
(x+x^2+x^3)=x*x*x+x*x*x^2+x*x*x^3+x*x^2*x^2+x*x^2*x^3+x*x^3*x^3+x^2*x^2*x^2+x^2*x^2*x^3+x^3*x^3*x^3=
=x^3+x^4+x^5+x^5+x^6+x^7+x^6+x^7+x^9=x^3+x^4+2x^5+2x^6+2x^7+x^9.

A simpler way to view this is to keep the product x^i*x^j*x^k if and
only if i(less or equal)<=j<=k

I forgot x^2*x^3*x^3=x^8 in the above result. An the ^3 int the left
hand.

.