Finding distinct integer solutions to $x_1 + x_2 + ...+ x_r = n$

Question

How many different (distinct $x_i$) non-negative integer solutions does the equation $x_1 + x_2 + ...+ x_r = n$ have?

We know that it has $n+r-1 \choose r-1$ non negative solutions. But how many are different?

actually I want to solve this problem... choose 5 card from 13 cards that every cards marked with number 1 to 13 . how many way sum of chosen cards if greater than 40. so i want to know how many distinct solution that equation has. because we have only one card from each one.

then i'm going to calculate this for n= 41 to 55

I assume we should read $r$ instead of $5$ and that $C(n,k)$ stands for the binomial coefficient $(n,k)$. — C. Falcon, Dec 16 '16 at 20:29
@C.Falcon: $C(n,k)$ is a fairly standard variant of $\binom{n}k$; $(n,k)$ is not. — Brian M. Scott, Dec 16 '16 at 20:30
@BrianM.Scott I recognize my point was not clear, I meant the binomial coefficient associate to the pair (or of index) $(n,k)$. — C. Falcon, Dec 16 '16 at 20:33
All the solutions are different. (Unless you mean distinct, $x_i\neq x_j$.) — Ribz, Dec 16 '16 at 21:29

score 1 · Answer 1 · edited Apr 13 '17 at 12:19

1

Hint: The question addressing the special case $r=13, 41\leq n \leq 65$ already has an answer here.

edited Apr 13 '17 at 12:19

Community

1

answered Dec 18 '16 at 15:43

Markus Scheuer

108,315

Fish NaCly · Answer 2 · 2023-12-19T09:21:11.470

use generating functions, consider integer $i$ can be chosen or not. If we choose it, it contributes $yx^i$ ; if not it contributes $1$. so the answer is

$$r![x^ny^r]\prod_{i=0}^n(1+yx^i)$$

use ferrers diagram we can prove that $$[y^r]\prod_{i=1}^n(1+yx^i)=\prod_{i=1}^r\frac{x^i}{1-x^i}$$

Here is my Mathematica code:

Coefficient[
 Series[Product[x^i/(1 - x^i), {i, 1, r}] + 
   Product[x^i/(1 - x^i), {i, 1, r - 1}], {x, 0, n}], x, n] r!

Finding distinct integer solutions to $x_1 + x_2 + ...+ x_r = n$

2 Answers2