My question is as follows: let there be $n$ different numbers $a_i$ in a set $A$, where each $a_i$ is a number between 0 and 1. How many different sets of values can I have that fulfill the condition $$\sum_{i=1}^{n} a_i=1$$ if the values of each $a_i$ may take on any of $k$ different values in the domain [0,1]? Does there exist a general formula to calculate this quantity? I have tried looking at formulas for numbers of combinations but run into trouble when trying to include the sum constraint.

(e.g. if allowed values of $a_i$ are $0, 0.25, 0.5, 0.75, 1$, then $k=5$).

Apologies in advance if I have used mathematical vocabulary inappropriately. Any advice on this matter would be much appreciated.

EDIT: values may be repeated

EDIT 2: Order does matter, e.g. {0.5, 0.5, 0} is distinct from {0.5,0,0.5}

  • $\begingroup$ In your example, suppose that $n=3$ and $k=5$. Is $0.5+0.5+0$ to be considered a different solution from $0.5+0+0>5$? $\endgroup$ – André Nicolas Jun 30 '15 at 4:30
  • $\begingroup$ In your second expression how is 0.5 > 5 ? $\endgroup$ – NewDogOldTricks Jun 30 '15 at 5:20
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    $\begingroup$ Sorry, typo, I meant $0.5+0+0.5$. In other words, does order matter? The intended $0.5$ turned into $0>5$, through the magic of the shift key. $\endgroup$ – André Nicolas Jun 30 '15 at 5:23
  • $\begingroup$ Ah yes, good question. Order does matter. $\endgroup$ – NewDogOldTricks Jun 30 '15 at 5:24
  • $\begingroup$ If you were looking for integer values for $a_i$, then the concept of "Integer Partitioning" would apply. However with $a_i$<1, I don't think there would be a closed form for the different number of ways. $\endgroup$ – NoChance Jun 30 '15 at 5:28

If you want lots of sets the values must be rational (maybe times one irrational). In your example, it would be equivalent to ask about subsets of $\{0,1,2,3,4\}$ that sum to $4$. In your case $4$ can be achieved $5$ ways: $1+1+1+1,1+1+2,1+3,2+2,4$. If you allow different orders it becomes $8$. If you allow $0$ there are infinitely many, as you can have as many zeros as you like. Not allowing $0$ these are the compositions of $n$. There are $2^{n-1}$ of them, which you can see by a stars and bars argument. Put $n$ stars in a row. There are $n-1$ places between them were you can choose to put a bar or not. Each set of bars gives a composition.

  • $\begingroup$ (1) yes, 0 is allowed. (2) values may be repeated. Would this change what you said above? $\endgroup$ – NewDogOldTricks Jun 30 '15 at 5:18

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