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I am really not mathematically inclined so whatever I'm asking might have already an obvious solution.

For most combination problems, we are given the choice to add our own values (is it n?), decide on the combination size, m, and generate the results

My question is, is it possible, if we add our number values (e.g. A:10, B:20, C:30) but instead of combination size (m), we have a choice to pick a total number combination, lets say Q, (Q= 40) and the algorithm show possible number values that could make up that total number value combination (Q).

In this case, the possible number value combinations could be

Q (Total Number Combination)= 40, where A:10, B:20, C:30

(1) A+A+A+A (10+10+10+10)

(2) B+B (20+20)

(3) A+A+B (10+10+20)

(4) A+C (10+30)

Could also allow for non-duplicates of values, in fact I would prefer that in a single combination to have no duplicates.

This is just a simplified example of what I'm doing, if you need more clarification, just buzz!

Thank you in advance Maths community! :)

N. F. Taussig
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San
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  • So you're asking for the number of solutions, in nonnegative integers, of $10x+20y+30z=40$. – Gerry Myerson Mar 14 '18 at 07:35
  • @GerryMyerson Thank you for your reply. I've read up on nonnegative integers briefly. Please correct me if I'm wrong, the solutions we can get from such equation is the number of integer solutions for the equation. Is it possible to know the possible combination does each of the 35 integer solutions entail? – San Mar 14 '18 at 08:35
  • I don't understand. What 35 integer solutions? 35 integer solutions of what equation? What does "combination" have to do with anything? – Gerry Myerson Mar 14 '18 at 12:04
  • @GerryMyerson Ooops, if the number of solutions to a particular nonnegative integers equation give rise to 35, is it possible to know what are the possible values (how many of x, y and z) that give rise to 35? Or am I understanding the concept of nonnegative integers all wrong? Let me read up on it abit more. – San Mar 15 '18 at 01:08
  • @GerryMyerson or perhaps I should just state what I'm doing now for work. Part of it involves exon calculations to know how many of such exons made up a particular band size. For example, Exon 2=39, Ex 3=147, Ex 4=136, Ex 5=98, Ex 6=68 A particular band size give rise to 341. I need to know what exon(s) give rise to that particular band. For this case its fairly easy to figure out, in which 341= Ex 2+4+5+6 Going back to the concept of nonnegative integers equation, if an equation give rise to 35 solutions, is it possible to know their possible combinations, as per the exon concept? – San Mar 15 '18 at 01:16
  • $a+b+c+d=4$ has 35 solutions, and you can certainly write them all out. Four of them have a 4 and four 0s; 12 of them have a 3, a 1, and two 0s; six of them have two 2s and two 0s; 12 of them have a 2, two 1s, and a 0; and one of them has four 1s. But you're getting close to some very difficult problems. In the subset sum problem, you're given positive integers $a_1,a_2,\dots,a_n$, and a target $T$, and you have to decide whether you can get $T$ as a sum of some of the $a_i$ (not using any $a_i$ more than once). For large $n$, no one knows a computationally feasible method. – Gerry Myerson Mar 15 '18 at 01:45
  • You may find https://math.stackexchange.com/questions/15521/making-change-for-a-dollar-and-other-number-partitioning-problems informative. – Gerry Myerson Mar 15 '18 at 01:51
  • Or http://www.jstor.org/stable/10.4169/194762110x489224 (Martin Erickson, Change for a dollar, change for a million, Math Horizons 17, No. 3 (February 2010) 22-25). – Gerry Myerson Mar 15 '18 at 02:00
  • @GerryMyerson Thank you so much for your inputs. I'll def read them up. Really useful. – San Mar 15 '18 at 02:21

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