Combinations of positive integer solutions to $x+y+z=200$ where $x,y,z \leq 100$ 
A piece of art receives an integer mark from $0$ to $100$ for each of
  the categories design, technique and originality. In how many ways is
  it possible to score a total mark of $200$?

I got the answer of $5151$ by writing out some cases. Is there a faster way to do this?
 A: Basically, we need to find out the number of solutions to $x_0+x_1+x_2=200$ with $0\leq x_i\leq 100$ for all $i$ and furthermore it should hold that $x_0+x_1\geq 100$
Once $x_0$ and $x_1$ are assigned, $x_2$ is automatically determined as $200-x_0-x_1$
Assigning $x_0=0$, there is only one choice for $x_1$ which is $100$
Assigning $x_0=1$, there are two choices for $x_1$, which are $99$ and $100$
Assigning $x_0=k$, there are $k+1$ choices for $x_1$, from $100-k$ to $100$
So, there are a total of $$\sum_{k=0}^{100}(k+1)=\sum_{k=1}^{101}k=\frac{101\times 102}2=5151$$ choices.
A: Since the marks can range from $0$ to $100$, we wish to solve the equation
$$x_1 + x_2 + x_3 = 200 \tag{1}$$
in the nonnegative integers subject to the restrictions that $x_1, x_2, x_3 \leq 100$.
If there were no restrictions, then a particular solution of equation 1 corresponds to the placement of two addition signs in a row of $200$ ones.
Let's illustrate with the equation $y_1 + y_2 + y_3 = 10$.
$$1 1 1 1 + 1 1 1 1 1 + 1$$
corresponds to the solution $y_1 = 4$, $y_2 = 5$, $y_3 = 1$, while 
$$1 1 1 1 1 + + 1 1 1 1 1$$
corresponds to the solution $y_1 = 5$, $y_2 = 0$, $y_3 = 5$.
The number of solutions of equation 1 in the nonnegative integers is 
$$\binom{200 + 3 - 1}{3 - 1} = \binom{202}{2}$$
since we must choose which two of the $202$ positions required for $200$ ones and two addition signs will be filled with addition signs.
From these, we must subtract those solutions in which one of the variables exceeds $100$.  Notice that at most one variable may exceed $100$ since $2 \cdot 101 = 202$.  There are three ways to choose the variable that exceeds $100$.  Suppose it is $x_1$.  Then $x_1' = x_1 - 101$ is a nonegative integer.  Substituting $x_1' + 101$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + 101 + x_2 + x_3 & = 200\\
x_1' + x_2 + x_3 & = 99 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with 
$$\binom{99 + 3 - 1}{3 - 1} = \binom{102}{2}$$
Hence, there are 
$$\binom{3}{1}\binom{101}{2}$$
solutions that violate the restrictions $x_1, x_2, x_3 \leq 100$.
Hence, the number of ways a total mark of $200$ can be obtained on three tests in which the maximum possible score is $100$ is 
$$\binom{202}{2} - \binom{3}{1}\binom{101}{2} = 5151$$
as you found.
