Desperate need, for help on Permutations!!

I've been searching all the web and math pages for a week now and i just can't find what im looking for.
How can i calculate all the possible permutations that add up to 100 using only whole numbers and using 10 integers.
Like such:

0 0 0 0 0 0 0 0 0 100
0 0 0 0 0 0 0 0 1 99
0 0 0 0 0 0 0 1 0 99

Please, there must be a way. And i need to finish up a project in a week and this is my last remaining obstacle.
I would really appreciate your help, thanks for your time

The problem is not very well defined.
For example, do you count 0 0 0 0 0 0 0 0 1 99 and 0 0 0 0 0 0 0 0 99 1 as two different ways?
Because you used the word permutation, they should be regarded as different.

Then the two 45 1 1 1 1 1 0 0 0 0 50 and 1 0 1 0 50 1 0 1 45 0 would be different.
If so, there is an easy answer to that problem: Combination of 109 things taken 9 at a time.

But if the problem means something else then the task of counting the outcomes could be quite difficult.

They are permutations, they are different!

PKA you are correct, 0 0 0 0 0 0 0 0 1 99 and 0 0 0 0 0 0 0 0 99 1, are different.
As are 45 1 1 1 1 1 0 0 0 0 50 and 1 0 1 0 50 1 0 1 45 0.
But i didn't understand your answer. I need all possible permutations, like the previous four just written. I don't need to know how many there are as much as i need them all, written like such.
So i can Implement them into a database program, (excell and such) and run particular tests.
Please, if you know how i may obtain such permutations, i would greatly appreciate it.

You have in effect this equation:
\(\displaystyle \L x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} = 100,\;0 \le x_k \le 100.\)

There is a well known formula for putting 100 identical objects, in this case 1’s, into ten different boxes, the x’s.
It is \(\displaystyle \L \begin{array}{l}
\left( {\begin{array}{c}
{100 + 10 - 1} \\
{10 - 1} \\
\end{array}} \right) = \frac{{(109)!}}{{\left( {9!} \right)\left( {100!} \right)}} = {\rm{4263421511271}} \\
\end{array}.\)

pka said:

It is \(\displaystyle \L \begin{array}{l}
\left( {\begin{array}{c}
{100 + 10 - 1} \\
{10 - 1} \\
\end{array}} \right) = \frac{{(109)!}}{{\left( {9!} \right)\left( {100!} \right)}} = {\rm{4263421511271}} \\
\end{array}.\)

Click to expand...

>I don't need to know how many there are as much as i need them all, written like such.

I suggest you get a good supply of pencils, Pablo :wink:

Thanksss

Hahaha, I should.
I guess I'll have to find a way around it then.
Thenk you very much pka. You've been of great help. I appreciate you taking your time to answer my question.

Pablo