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)
 A: Plain Stars and Bars does  not quite work here. The integers are supposed to be distinct, and plain Stars and Bars pays no attention to that. (It can be adjusted.)  
But we will not use it. We count the triples $(a,b,c)$ such that $a\lt b\lt c$, then multiply by $3!$.  
Adding $47$ sounds like a good idea; but it turns out that the symmetry of $\pm 47$ is very useful.
There are $95$ numbers in our list. Suppose that we choose $3$ of them. Some choices add up to $0$. Some choices have positive sum. Some have negative sum. By symmetry, just as many have positive sum  as negative sum. 
So if we find the number $z$ of choices with sum $0$, we will be finished. For let $p$ be the number with positive sum. We have $z+2p=\binom{95}{3}$. We solve for $p$, and find that our answer is
$$3!\frac{\binom{95}{3}-z}{2}.$$
Now we only want to find $z$, the number of unordered triples, or equivalently triples with $a\lt b\lt c$, such that $a$, $b$, and $c$ satisfy the size constraints, and $a+b+c =0$. Much simpler!
Temporarily, we leave it to you to find $z$. That is very doable.
Finding $z$: We concentrate on the middle term. If this is $0$, there are $47$ ways to complete the triple. Now look at middle term $1,2,3, \dots$. (There are just as many with middle terms $-1, -2,-3,\dots$.) 
If the middle term is $1$, the big term can be any of $2$ to $46$, so $45$ possibilities, and now the small term is determined. If the middle term is $2$, there are $43$ possibilities, and so on until when the middle term is $23$, there is only $1$ possibility for the big term, namely $24$. So the total with positive middle term is $1+3+5+\cdots +45$, which is $23^2$.  It follows that
$$z=47+(2)(23^2),$$
and we are finished.
Remark: Precisely the same reasoning works if we replace the bound $\pm 47$ by $\pm B$.  One can also deal with intervals that are less overtly symmetric. But the symmetry or near symmetry of the condition  $a+b+c$ is crucial for an argument of the above type.  
A: There’s a better way: introduce $d_1=282-(a_1+b_1+c_1)$ and count solutions to
$$a_1+b_1+c_1+d_1=282$$
subject to the known constraints on $a_1,b_1$, and $c_1$ and the constraint $0\le d_1\le 140$.
A: 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: 
$$\begin{align}
\mathscr{N}^{distinct}_{a+b+c>0} &= \mathscr{N}_{a+b+c>0} - \frac12\left( (6N+1)(2N+1) - 6\lfloor\frac{N}{2}\rfloor - 1\right)\\
&= \frac12 (N-1)N(8N+5) + 3\lfloor\frac{N}{2}\rfloor\\
&= 411930
\end{align}$$
A: We argue about the set $A_n$ of admissible lattice points in the "cube" $[-n,n]^3\cap{\mathbb Z}^3$. Note that $A_1=\emptyset$ and $A_{n-1}\subset A_n$ for $n\geq2$. Denote the number of points in $A_n$ by $f(n)$.
We begin with $f(2)$. The set $\{-2,-1,0,1,2\}$ contains four triples of different numbers  with a positive sum, namely $\{2,1,0\}$, $\{2,1,-1\}$, $\{2,1,-2\}$, $\{2,0,-1\}$. It follows that $f(2)=24$.
We now count the number of points in the set $\Delta_n:=A_n\setminus A_{n-1}$. Draw a cube, and you will see that $\Delta_n$ consists of three "quadratic plates" from which a triangle has been taken away, and three little triangles of exactly the same size. Doing a careful count with the plate $c=n$ and the little triangle on the face $c=-n$ one obtains
$$\#\Delta_n=\cases{3(4n^2-5n+2)\quad &($n$ even) \cr
3(4n^2-5n+1)&($n$ odd)\cr}\ .\tag{1}$$
The distinction between even and odd $n$ stems from the diagonal $a=b$ that has to be removed: If $n$ is even there is a point with $a+b=-n$ on this diagonal, and when $n$ is odd there is no such point.
Formula $(1)$ can be rewritten as
$$\#\Delta_n=12n^2 -15n +{9\over2}+{3\over2}(-1)^n\qquad(n\geq3)\ .$$
It follows that
$$f(n)=24+\sum_{k=3}^n\left(12n^2-15n+{9\over2}+{3\over2}(-1)^n\right)\ ,$$
and this evaluates to
$$f(n)={1\over4}\bigl(16 n^3-6n^2-4n-3+3(-1)^n\bigr)\qquad(n\geq2)\ .\tag{2}$$
Note that we even get $f(1)=0$, though in this case the counting might be considered somewhat fishy. (I have to admit that I also did a brute force count with Mathematica and  up to $n=47$ obtained the numbers produced by $(2)$.) 
In particular $f(47)=411930$, in accordance with @André Nicolas' result (when unpacked) for this case.
