Finding the number of triples $(a,b,c)$ with $-47 \leq a,b,c \leq 47$ and $a+b+c >0$ 
Kate is looking for ordered triples $(a,b,c)$ of distinct integers such that $-47 \leq \; a,b,c \; \leq 47$ and $a+b+c>0$. How many such ordered triples can Kate find?

My first step was to let  $a_1=a+47, \; b_1=b+47$ and $c_1=c+47$ so that $a_1+b_1+c_1>141$ and  $0 \leq \; a,b,c \; \leq 94$. Do I simply have to solve for all $a_1+b_1+c_1=142, \; 143, \cdots, 282$ and sum these results. Or is there a better way? 
You can solve the above sub problems using a stars and bars method http://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

Let $N = 47$ and $X = \{-N,\ldots,N\}$. We have $|X^3| = |X|^3 = (2N+1)^3$.
Let $\mathscr{N}_{???}$ be the number of solutions for any equation ??? involving $(a,b,c) \in X^3$.
Because of symmetry, 
$$\mathscr{N}_{a+b+c>0} = \mathscr{N}_{a+b+c<0} \implies \mathscr{N}_{a+b+c>0} = \frac12 \left((2N+1)^3 - \mathscr{N}_{a+b+c=0} \right)$$ 
For any fixed $c$, the number of solutions for $a + b + c = 0$ is $(2N+1) - |c|$, we have:
$$\mathscr{N}_{a+b+c=0} = \sum_{c = -N}^{N} \left( (2N+1) - |c|\,\right) = (2N+1)^2 -2\frac{N(N+1)}{2} = 3N^2+3N+1$$
and hence
$$\mathscr{N}_{a+b+c>0} = \frac12\left((2N+1)^3 - (3N^2+3N+1)\right) = \frac12 N (8N^2 + 9N + 3) = 425303$$
Method 2
Another way to get the same answer is try to attack a more general question:

Given $n \in \mathbb{Z}$ and $d, L \in \mathbb{Z}_{+}$, what is the
  number of integer solutions for $a_1 + a_2 + \cdots + a_d = n$
  where all $ 0 \le a_i < L$.

Let $\varphi_{d}(n)$ and $\varphi_{d}(n;L)$ be the number of non-negative integer solutions  where $\sum_{i=1}^{d} a_i = n$ and those with further restriction that $a_1,\ldots,a_d < L$. i.e.
$$\begin{align}
\varphi_{d}(n) 
&= \left|\,\{ (a_i) \in \mathbb{N}^{d} : \sum_{i=1}^{d} a_i = n \}\,\right|\\
\varphi_{d}(n;L)
&= \left|\,\{ (a_i) \in \mathbb{N}^{d} : \sum_{i=1}^{d} a_i = n \wedge a_1, \ldots a_d < L \}\,\right|
\end{align}$$
By induction, it is not hard to show:
$$\varphi_{d}(n) = \begin{cases}\binom{n+d-1}{d} = \frac{n(n+1)\cdots(n+d-1)}{d!},& n \ge 0\\ \\0, & n < 0\end{cases}$$
To compute $\varphi_{d}(n;L)$, we can start with the set of solutions for $\varphi_{d}(n)$
and then proceed to remove those solutions with at least one $x_i \ge L$. For first approximation, we get:
$$\begin{align}\varphi_{d}(n;L) 
&= \varphi_{d}(n) - \sum_{j=1}^{d} \left|\{ (a_i) \in \mathbb{Z}^d : \sum_{i=1}^{d} a_i = n \wedge a_j \ge L \}\right| + \cdots\\
&= \varphi_{d}(n) - d \varphi_{d}(n-L) + \cdots\\
\end{align}$$
However, for those solutions with at least two $a_i \ge L$, this procedure removed it more than once. We need to add back their contribution. For second approximation, we get:
$$\begin{align}\varphi_{d}(n;L)
&= \varphi_{d}(n) - d \varphi_{d}(n-L) + \sum_{1 \le j < k \le d} \left|\{(a_i) \in \mathbb{Z}^d : \sum_{i=1}^{d} a_i = n \wedge a_j, a_k \ge L \}\right| + \cdots\\
&= \varphi_{d}(n) - d \varphi_{d}(n-L) + \binom{d}{2} \varphi_{d}(n-2L) + \cdots
\end{align}$$
For those solutions with at least three $a_i \ge L$, this again add back too much. We can use Inclusion-Exclusion to organize this mess. In the end, we will have:
$$\varphi_{d}(n;L) 
= \sum_{k=0}^{d} (-1)^k \binom{d}{k} \varphi_{d}(n - kL)
= \sum_{k=0}^{\min(d,\lfloor\frac{n}{L}\rfloor)}(-1)^k \binom{d}{k} \varphi_{d}(n-kL)
$$
Back to our original problem, it is clear the number of solutions for 
$$a + b + c < 0,\quad|a|, |b|, |c| \le N$$
is equal to the number of solutions for $$x + y + z = 3N,\quad 0 \le x, y, z < 2N+1$$
Since $\lfloor\frac{3N}{2N+1}\rfloor = 1$, we get:
$$\begin{align}\mathscr{N}_{a+b+c>0}
&= \mathscr{N}_{a+b+c<0}\\
&= \varphi_3(3N;2N+1)\\
&= \varphi_3(3N) - \binom{3}{1}\varphi_3(3N-(2N+1))\\
&= \varphi_3(3N) - 3\varphi_3(N-1)\\
&= \frac12 N(3N+1)(3N+2)) - \frac12 (N-1)N(N+1)\\
&= \frac12 N(8N^2+9N+3)\\
&= 425303
\end{align}$$
UPDATE
Oops, forget the requirement that $a, b, c$ are distinct. We need to count the
cases where $a, b, c$ fails to be distinct. There are two possible cases:


*

*Case 1, two of $a, b, c$ is the same. There are 3 possible subcases.
For the subcase where $a = b \ne c$, we have $2N(2N+1)$ combination. 
$2 \lfloor\frac{N}{2}\rfloor$ of which sums to $0$.
The other 2 subcases give same amount of counting.

*Case 2. $a = b = c$. There are $2N+1$ combination, only 1 of them sums to $0$.


From this, the number of solutions for $a + b + c > 0$ with $a \ne b \ne c$ is given by: 
$$\begin{align}
\mathscr{N}^{distinct}_{a+b+c>0} &= \mathscr{N}_{a+b+c>0} - \frac12\left( (6N+1)(2N+1) - 6\lfloor\frac{N}{2}\rfloor - 1\right)\\
&= \frac12 (N-1)N(8N+5) + 3\lfloor\frac{N}{2}\rfloor\\
&= 411930
\end{align}$$
