1. ## How many solutions

How many solutions does the equation

$x+y+z=100$

have on the set of positive integers?

2. Originally Posted by perash
How many solutions does the equation

$x+y+z=100$

have on the set of positive integers?
First of all transform the equation to a+b+c = 97.
Now the solution to this is (97+3-1) choose (3-1) = 99 choose 2

3. Originally Posted by perash
How many solutions does the equation

$x+y+z=100$

have on the set of positive integers?
The equation $x_1+x_2+...+x_k = n$ can be inverstigated in the following way. First construct a row of $n$ $1$'s: $1 \ 1 \ 1 ... \ 1$.
Now between any space you can place a break $\bold{b}$ and that will give you a possible $(x_1,...,x_k)$ pair.
For example, to illustrate, $n=5$, $k=3$, we can break it as $1 \ 1 \ \bold{b} \ 1 \ \bold{b} \ 1 \ 1$. That gives us $x_1=2, x_2=1,x_3=2$. Thus, given $n$ as an integer we place $k-1$ breaks in between the $n-1$ breaks. There are a total of ${{n-1}\choose{k-1}}$ ways of doing this.

@Isomorphism: The proof of the combinatorial formula you used rests on what I did.

4. Originally Posted by ThePerfectHacker
@Isomorphism: The proof of the combinatorial formula you used rests on what I did.
I know that
But I was too lazy to write that argument out. Isn't there another way too look at it-The coefficient of polynomial way?

5. Hello, perash!

How many solutions does the equation: . $x+y+z=100$
have on the set of positive integers?
Here's a "visual" explanation of Isomorphism's solution . . .

We have a 100-inch "yardstick", marked in inches.

We will cut it into three pieces by making two cuts.

There are 99 inch-marks and we will chose two at which to cut.

Answer: . ${99\choose2} \:=\:4851$ ways.