100% shared over 17 If you have 100% and you need to share it over 17 things and every thing should get atleast 1% how do many different ways are there to do this?
I'm a highschool student and was trying to write a story in which someone simulates how she has 17 ingedients and simulates every possiblle way to do this if every ingedient should get atleast 1%. I'm personally not very advanced in maths but did try to solve it atleast partly. 
I realised that I could never solve this alone so I asked some friends
What we got was that we could simplify the equation by removing the 1% by doing 100-17=83. Now we had to share this 83 over 17 whilst using all of it. Niether of us knew how to do this so we simplified it to 5 cookies and 2 people then 3 people, 4-,5-. We did the gind to work out all of these trying to find a pattern but we didnt succeed. We know that it has something to do with combinatorics.
My question is if someone can give us Hints or fomules so we can try to solve it ourselves.
My apologies if my Grammar isn't correct, I'm not a Native English speaker and still struggle with the language sometimes.
 A: Let us simplify the problem : 
Let us assume that you have $3$ persons and $10$ euros to share between them, each person receiving at least $1$ euro.
As you have well seen it, this issue is equivalent to share $7$ euros, some persons hopefuly receiving nothing (knowing that at the end, one euro will be added to the "bag" of each person).
The "stars and bars method" gives an easy way to solve such a problem.
Here is how : to each possible money distribution, we associate in a bijective way a binary representation with 7 zeros (say they represent 1 euro coins...) and 2 ones (the bars) acting as barriers between zeros like in these examples :
$$\begin{cases}O|OO|OOOO \ & \leftrightarrow  \ & \text{Alice has 1 euro, Bob has 2 euros, Charles has 4 euros}\\
OOOOO|OO| \  & \leftrightarrow  \ & \text{Alice has 5 euros, Bob has 2 euros, Charles has nothing}\\
||OOOOOOO \ & \leftrightarrow  \ & \text{Alice has nothing, Bob has nothing, Charles has 7 euros}\end{cases}.$$
Please note that we always take the persons in the same order : Alice, then Bob, then Charles.
We are now with a classical combinatorics issue : counting how many binary sequences (i.e., with zeros and ones) with n digits contain exactly $p$ digits "ones".
This number is the binomial coefficient : $\binom{n}{p}=\binom{7+2}{2}=\dfrac{9!}{7!2!}=\dfrac{9 \times 8}{2}=36.$
Can you proceed from here ?
