Different ordered triples $(a,b,c)$ of non-negative integers

Question

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?

Answer 1

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.

Answer 2

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;].

Answer 3

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

$$ \sum_{a=0}^{50} (51-a) = 51\cdot 51 - \frac{50\cdot 51}{2} = 26\cdot 51 = 1326 $$

Answer 4

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-

$\dbinom{n+r-1}{r-1}$

Answer 5

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.