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This must be basic, but my Mathematica (and overall programming) skill level is too low to find a handy solution:

I need to generate all possible lists of $m$ integers (I call these integers $x_i$'s, where $i=1, \dots, m$, with $0 \leq m \leq X$) such that $$\sum_{i=1}^m x_i = X.$$

For each list generated, I then need to count how many of the $x_i$'s are $0$'s, $1$'s, etc., up to $X$.

I can tell that these $m$ integers will be arranged in the following possible configurations:

$$ \frac{(m+X-1)!}{(m-1)! \ X!}.$$

Also, that for a given $x_i$, there will be the following number of possible configurations: $$\frac{(N+W-x_i-2)!}{(N-2)! (W-x_i)!}.$$

Say, $m=3$ and $X=3$. Then the possible configurations are $$\{3, 0, 0\},$$ $$\{2, 1, 0\},$$ $$\{2, 0, 1\},$$ $$\{1,2,0\},$$ $$\{1,0,2\},$$ $$\{1,1,1\},$$ $$\{0,3,0\},$$ $$\{0,2,1\},$$ $$\{0,1,2\},$$ $$\{0,0,3\}.$$

So, here, there is one 3 and two 0's in the first list, etc.

Now, I need Mathematica to help me generate these lists (10 in my example), but for arbitrary (though not very large) values of $m$ and $X$.

Again, my next step is to make Mathematica take each list in turn and count the $x_i$'s that are zeroes, ones, etc. up to $X$. But that I think I can try and figure out myself.

I did look up similar questions in the archive but couldn't find a pointer to the solution.

Julio Huato
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3 Answers3

4

You can use IntegerPartitions and Permutations for this:

partitions[m_, X_] := Catenate[
    Permutations /@ IntegerPartitions[X, {m}, Range[0,X]]
]

For your example:

partitions[3, 3]

{{3, 0, 0}, {0, 3, 0}, {0, 0, 3}, {2, 1, 0}, {2, 0, 1}, {1, 2, 0}, {1, 0, 2}, {0, 2, 1}, {0, 1, 2}, {1, 1, 1}}

Carl Woll
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0

You can simply Select from the generated set of Tuples:

x = 3; m = 3;
lists = Select[Tuples[Range[0, m], m], Total@# == x &];

This will give you your lists:

lists // Grid

0   0   3
0   1   2
0   2   1
0   3   0
1   0   2
1   1   1
1   2   0
2   0   1
2   1   0
3   0   0

To count the occurrences of the digits, just use Tally:

Tally /@ lists // Grid

{0,2}   {3,1}   
{0,1}   {1,1}   {2,1}
{0,1}   {2,1}   {1,1}
{0,2}   {3,1}   
{1,1}   {0,1}   {2,1}
{1,3}       
{1,1}   {2,1}   {0,1}
{2,1}   {0,1}   {1,1}
{2,1}   {1,1}   {0,1}
{3,1}   {0,2}

The first line means: "0 occurs 2 times and 3 occurs once". The next line has three different digits...

For longer lists or different Totals, just change x or m in the first line.

Matthias Schwarze
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0

A bit slow but this might work:

soln1[m_, X_] := 
Module[{}, n = (m + X - 1)!/(m - 1)!/X!; 
FindInstance[
Sum[x[i], {i, 1, m}] == X && And @@ Thread[Array[x, m] >= 0], 
Array[x, m], Integers, n]]

Array[x, 3] /. soln1[3, 3]

(* {{0, 0, 3}, {0, 1, 2}, {0, 2, 1}, {0, 3, 0}, {1, 0, 2}, {1, 1, 1}, {1, 2, 0}, {2, 0, 1}, {2, 1, 0}, {3, 0, 0}} *)

user11946
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