Combinatorics Problem: N quanta condensation into s states under occupancy limits

[combodummy]

Hello! I am a chemist struggling with finding solutions to problems I pose to myself; I am hopeful someone here can help me or at least steer me to other possible approaches.

Anyway, here is a specific simplified problem representative of my chemometric interests.

In how many distinct ways, say W, can 100 chips be distributed among 5 bowls such that the sum of the respective chip counts is 100, when it is further specified that

a) the chip count in bowl 1 is C, C being any integer from 21 to 100 inclusively, and

b) the balance of chips, for given C, is distributed randomly among the 4 remaining bowls such that none of the resulting chip counts exceed C-1.

Corollary Question: if after random distribution of the balance of chips as above, the bowls 2- 4 are rank ordered left to right from highest to lowest chip count (zeroes and ties allowed), how many distinct 4-tuple partitions (say P) are produced and what are their respective degeneracies (g sub p)so that the sum of the degeneracies over all P 4-tuples is equal to W.

Note: in terms of states, W represents the total number of micro-states available to the N quanta, whereas P represents the number of distinct macro-states with respective degeneracies g sub p.

Thanks in advance, Combodummy.

 

[daon2]

I don't understand your question. You have 100 "chips" distributed into 5 bowls. It would be impossible to get a 100 chip count in each bowl.

 

[Deleted member 4993]

I don't understand your question. You have 100 "chips" distributed into 5 bowls. It would be impossible to get a 100 chip count in each bowl.

I think he wants to divide 100 chips in five bowls according to the forllowing rule:

C + A + B + D + E = 100 and 20<C<101 and (C-1) ≥ A, B, D & E

 

[daon2]

In that case, it is a question similar to this one:

http://www.freemathhelp.com/forum/threads/76279-Enumeration-found-answer-but-need-quick-help

You need the number of 4-partitions of (100-C) where each bowl(digit) does not exceed (C-1). Perhaps use Pka's generating function as a guide.

 

[combodummy]

In that case, it is a question similar to this one:

http://www.freemathhelp.com/forum/threads/76279-Enumeration-found-answer-but-need-quick-help

You need the number of 4-partitions of (100-C) where each bowl(digit) does not exceed (C-1). Perhaps use Pka's generating function as a guide.

I am totally unfamiliar with modern math/computer program jargon and/or capabilities. What is Pka's generating function and where might I find how to use it? Thanks.

 

[daon2]

I am totally unfamiliar with modern math/computer program jargon and/or capabilities. What is Pka's generating function and where might I find how to use it? Thanks.

I was referring to a post user Pka made in the thread I linked.

If I understood it correctly (please correct me if I am mistaken), for a given C, I believe the coefficient of \(\displaystyle x^{100-C}\) in the polynomial \(\displaystyle \displaystyle \left (\sum_{n=0}^{C-1} x^n\right)^{4}\) will be the result you want.

For example, C=21 gives the coefficient for x^79 as 4, which makes sense:

21 - 20 - 20 - 20 -19
21 - 20 - 20 - 19 -20
21 - 20 - 19 - 20 -20
21 - 19 - 20 - 20 -20

These are the only ways to have 100 chips if: The first bowl has 21 chips, and no other bowl can have more than 20 chips.

 

[mmm4444bot]

May some bowls be empty, after the chip distribution?

 

[Deleted member 4993]

Since

a) the chip count in bowl 1 is C, C being any integer from 21 to 100 inclusively,

A, B, D and E can be zero.

 

[mmm4444bot]

Exactly.

 

[combodummy]

Thanks.

I was referring to a post user Pka made in the thread I linked.


If I understood it correctly (please correct me if I am mistaken), for a given C, I believe the coefficient of \(\displaystyle x^{100-C}\) in the polynomial \(\displaystyle \displaystyle \left (\sum_{n=0}^{C-1} x^n\right)^{4}\) will be the result you want.

For example, C=21 gives the coefficient for x^79 as 4, which makes sense:

21 - 20 - 20 - 20 -19
21 - 20 - 20 - 19 -20
21 - 20 - 19 - 20 -20
21 - 19 - 20 - 20 -20

These are the only ways to have 100 chips if: The first bowl has 21 chips, and no other bowl can have more than 20 chips.

Thank you Daon! I think I get the general idea now, and I understand what you said about the coefficient of X^(100-c). Also I agree with the four solutions you wrote out for c=21, these 4 solutions collapse to only one after rank ordering i.e. 21-20-20-20-19. How did you evaluate the various coefficient terms in that expansion.

 

[combodummy]

Thanks mmm444bot

May some bowls be empty, after the chip distribution?

Thanks . yes some bowls may be empty; say c =51 then (51, 49, 0,0, 0) would be a permissible solution, as would be the solutions generated by permuting the four bowls b2-b4.

 

[mmm4444bot]

Thanks. Got it. (I was trying to understand Daon's comment.)

 

[daon2]

Thank you Daon! I think I get the general idea now, and I understand what you said about the coefficient of X^(100-c). Also I agree with the four solutions you wrote out for c=21, these 4 solutions collapse to only one after rank ordering i.e. 21-20-20-20-19. How did you evaluate the various coefficient terms in that expansion.

I used wolfram alpha.

http://www.wolframalpha.com/input/?i=+Coefficient((sum_{k=0}^{20}x^k)^4,x,79)

I'll note that the coefficients in the expansion are symmetric. So choosing C=21 the coefficients of x and x^79 will give the same answer.