Different ordered triples $(a,b,c)$ of non-negative integers How many different ordered triples $(a,b,c)$ of non-negative integers are there such that $a+b+c=50$?
I tried to list the possibilities but the list is way too long, I know how to find the ordered doubles $(x,y)$ such that $x+y=50$, I just have to list them like:
$(0,50)\\(1,49)\\ \vdots\\(49,1)\\(50,0)$
Which is just simply $(50-0)+1=51$.  But this is too long to count.  I suppose there's a better way to do this?
 A: Let's try to solve this problem starting from where you're at. 
If the question's asking for different ordered doubles such that $x+y=50$ then there're $51$ different ordered pairs.  So, the equation that $x+y=n$ can be satisfied by $n+1$ different ordered pairs $(x,y)$:
$$(0,n),(1,n-1),\ldots ,(n-1,1),(n,0)$$
For your case, asking for different ordered triples, we can partition the solution of $a+b+c=50$ into disjoint cases in which $c=0,c=1,\ldots ,c=49,c=50$.  
So, if $c=0$, then $a+b=50\implies $ there are $50+1=51$ possible answers for $(a,b,0)$
Now, for $c=1\implies a+b=49\implies $ there are $49+1=50$ possible answers for $(a,b,1)$
$$\vdots$$
So, the general trend is that for each $c=m \implies$ that there are $50-m+1$ ways to choose $a$ and $b$ for that specific $c$.
$\therefore$ There are $51+50+\ldots +3+2+1=\frac{51\times 52}{2}=1326$ different ordered triples.
A: Any such triple can be encoded as a binary sequence of length $52$, containing exactly $50$ zeros and two separating ones, written as $|\>$: The number of zeros to the left of the first $|$ is $a$, the number of zeros between the two $|$'s is $b$, and the number of zeros to the right of the second $|$ is $c$.
There are ${52\choose 2}=1326$ such sequences.
A: BEGGAR'S METHOD:
Let's say we have 50 identical coins and we have to distribute it into 3 beggers.
Similar to $$a+b+c=50$$ We can solve this problem by combinations and permutations.[I try these types of questions like this]
Try it!

No. of ways is $C_{2}^{52}=51\times 52/2=1326$. that can be just seen as making 2 lines to divide 50 coins placed in a row into 3 parts. So, there are 52 places to draw a line from which we have to chose 2,[52 are including the ends which will create a empty part or solution for one variable=0;].

A: Let's make this simple.
Given a and b, where $a+b\leq 50$ and $a,b\geq 0$, $c$ is immediately defined to be $50-a-b$, and thus there's exactly one such triple for any given $a$ and $b$.
So the question is, how many combinations of $a$ and $b$ exist under those restrictions. So we work it out. If $a=50$, then $b=0$ is the only option. If $a=49$, then $b=0$ or $b=1$. For any particular value of $a$, you have that $b$ is limited by $0\leq b\leq 50-a$. So the total number of values of $b$ possible is $51-a$. And $0\leq a\leq 50$. So the number of ordered triples is
$$
\sum_{a=0}^{50} (51-a) = 51\cdot 51 - \frac{50\cdot 51}{2} = 26\cdot 51 = 1326
$$
A: Another way to do this is to use generating functions!  Since there are 51 different digits one can choose from (0 through 50) in any of the given positions, we could think of this as the polynomial sum(x^n, n=0...50).  Since we are choosing three of them, then we expand (using Maple/Wolfram Alpha), [sum(x^n, n=0...50)]^3, and look for the coefficient of the 50th term.  That coefficient is 1326!


A: There is yet another method to do this problem.
Let $x=a+1,y=b+1$ and $z=c+1$
Write down 53 as-
$1+1+1+\dots +1$
Now removing any 2 '+' signs and replacing with a comma will give us 3 integers x,y,z such that:
$x+y+z=53$ and $a+b+c = x-1+y-1+z-1=50$
Number of ways to remove 2 '+' signs out of 52 '+' signs = $\dbinom{52}{2}=1326$ 
On generalizing we find that number of solutions to $(x_{1},x_{2},\dots,x_{r})$ such that:
$x_{1}+x_{2}+\dots +x_{r}=n$ with $x_{i}>0$           $,\forall i:0\leq i\leq r$ is-
$\dbinom{n+r-1}{r-1}$
