How many $3$ member subsets $\{x, y, z\}$ of positive natural numbers have the sum $x + y + z = 100$? I have a math homework problem where I think I have to use Permutation/Combination.
The question is: How many $3$ member subsets $\{x, y, z\}$ of positive natural numbers have the sum $x + y + z = 100$?
I am trying to use the stars and bars method, so there would be $2$ bars ($1$ less than the number of variables) and $100$ stars. 
So is my final answer simply $\binom{102}{100} = 5151$?
Feedback would be appreciated :):)
Also, if this is correct, how would I take my answer a step further and find a solution where I don't include zero?
 A: Since you already know the more general stars and bars formula, there is a simple way to convert it to that form.
To force positive integers, set $X = x-1, Y = y-1, Z = z-1,$ so now you have
$X + Y + Z = 97$ for non-negative values of $X,Y,Z$
and the formula yields $\dbinom{99}{97} or \dbinom{99}{2}$ according to your preference
A: Here order of the summation is not counted since we need the number of sets of three numbers with sum equal to 100. That is, $10+70+20$ and $10+20+70$ both lead to the same set $\{10,20,70\}$. Hence we need to count the 3 partitions of 100. We can use the recurrence
$$P(n,k) = P(n-k,k)+P(n-1,k-1)$$ where $P(n,k)$ denotes the $k$ partition of $n$. We need $P(100,3)$. Using the above recurrence and the facts that $P(2m,2) = P(2m+1,2) = m$, it is seen that $P(100,3) = 833$
A: As you have used star and bar method, you have $100$ chocolates and you have to distribute them to $3$ students such that each student gets at least $1$ chocolate.
So we arrange $100$ chocolates in a row and there is $1$ gap in between $2$ chocolates.
So there is a total of $99$ gaps, and we have to use separators in two of these gaps, which can be done by $\displaystyle \binom{99}{2}$ ways.
A: Although you answer is almost true, but in the statement of your question is stressed count only $3$ member subset $\{x, y, z\}$, but as you see, you have count $1 + 1 + 98 = 100$ and etc. I think that you should consider this situations. So you should count the number of solutions of equation $x+2y = 100$, then you can suppose that $x = 2k$, and solve the equation $k + y = 50$.
