# How many combinations of certain variables add up to at least n?

I have a set of numbers, let's say: a=16 b=19 c=19 d=31 e=5 f=... etc.

How to create a simple function that shows how many combinations of the letters add up to at least the number n?

I have no clue what Mathematica Functions I should use to get started. Any help would be appreciated.

• As a starting point, you could use Tuple to generate all combinations of letters of a given length; however some combinaison would be counted twice (e.g. $(a,b)$ and $(b,a)$). – anderstood Mar 16 '17 at 0:23
• "At least"?? So you're interested in collections that might give a higher number too? That unbounds your search. – David G. Stork Mar 16 '17 at 0:54
• @DavidG.Stork Yes; at least indeed... but the size of the list is not unlimited... – GambitSquared Mar 16 '17 at 0:55
• So you DO want to allow higher numbers! OK. – David G. Stork Mar 16 '17 at 0:56
• Possible duplicate mathematica.stackexchange.com/questions/22397/… – Anjan Kumar Mar 16 '17 at 1:09

n=80

number of subsets

Length@Select[Total /@ Subsets[{13, 15, 19, 19, 23, 31}], # >= 80 &]


and if you want to see the selected subsets type

Select[Subsets[{13, 15, 19, 19, 23, 31}], Total[#] >= 80 &]

• The question states "at least", so you should use $>$ (not $<$). But I think the poser is mistaken in what he wants. – David G. Stork Mar 16 '17 at 0:55
• you are right! fixed – J42161217 Mar 16 '17 at 0:57
• Great! This is almost what I need! Now, how do I visualize the lists that actually fall within the right category? And doesn't this solution also allow for duplicates? – GambitSquared Mar 16 '17 at 1:11
• ListPlot[DeleteDuplicates[ Select[Subsets[{13, 15, 19, 19, 23, 31}], Total[#] >= 80 &]]] – J42161217 Mar 16 '17 at 1:16
• @Jenny_mathy This is the endresult: Length@Select[Subsets[{31, 9, 19, 14, 19, 19, 6, 16, 3, 5, 4, 3, 2}], Total[#] > 75&] It's the amount of possible coalitions after the Dutch elections tonight. – GambitSquared Mar 16 '17 at 1:43

First create a list of your allowable component integers:

myset = {1, 2, 3, 5, 20};


Then:

Length[Select[IntegerPartitions[20], ContainsOnly[myset]]]


(* 92 *)

IntegerPartitions gives a list of all the says you can add integers to give the candidate number. Select from all those possible lists the ones that contain only the elements in your allowable component integers using ContainsOnly (version 10). Then count (Length) that list.

If you must, you can kludge a function using set operations to select lists whose union with myset is myset.

• ContainsOnly? I don't have that function...? – GambitSquared Mar 16 '17 at 0:50
• @David G. Stork You can directly use IntegerPartitions[20, All, {1, 2, 3, 5, 20}] // Length, instead of using Select. – Anjan Kumar Mar 16 '17 at 1:04
• @AnjanKumar: Ah yes... I didn't know that IntegerPartitions had that option. Thanks. – David G. Stork Mar 16 '17 at 1:12