How many different ordered triples are there such that $a+b+c = 50$ and $a\geq b\geq c\geq 0$? What if $a>b>c>0$?

Question

How many different ordered triples are there such that $a+b+c = 50$ and $a\geq b\geq c\geq 0$? What if $a>b>c>0$? a,b, and c are all integers

I found this question from a textbook, but the author didn't give the answer so I don't know if my answers are correct, can anyone help me with this problem?

My answers:

if $a\geq b\geq c\geq 0$: $221$ possible triples

if $a>b>c>0$: $196$ possible triples

Thanks!

Answer 1

REVISITED BY CHANCE AND A SIMPLE ANSWER OFFERED

PART 1

All three numbers obviously can't be the same.
Of the $\binom{52}2 = 1326$ solutions given by stars and bars,
there will be $26\;\;$ with $\;\;2-1\;of\; a\; kind:\; 0-0-50\;\; to\;\; 25-25-0$
each with $3$ permutations So distinct triples with $6$ permutations each $= 1326 - 3*26 = 1248$
and final answer $= \frac{1248}6+26 =\boxed{234}$

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PART TWO

This time, we want all three digits to be different.
There will be a total of $\binom{49}2 = 1176$ arrangements
of which this time $24$ will be $2-1\;\;of\;a\;kind$
and final answer = $(1176 -24*3)/6 = \boxed{184}$

Answer 2

Using generating functions:

First rewrite the problem by using the substitutions $$d = a-b, \qquad e = b-c, \qquad f = c.$$ so we have $d,e,f \geq 0$ are integers such that $d + 2e + 3f = 50$.

The number of such triplets can be found as the coefficient of $x^{50}$ in the generating function $$\frac{1}{(1-x)(1-x^2)(1-x^3)}$$

By using partial fractions, we may find, $$\begin{align*}\frac{1}{(1-x)(1-x^2)(1-x^3)} &= \frac{17x^2-52x+47}{72(1-x)^3} + \frac{1}{8(1+x)} + \frac{1}{3(1-x^3)} - \frac{1}{9(1-x)} \\ &= \frac{17x^2-52x+47}{72}\sum_{n=0}^\infty \binom{n+2}{2}x^n + \frac{1}{8}\sum_{n=0}^\infty (-1)^nx^n + \frac{1}{3}\sum_{n=0}^\infty x^{3n} - \frac{1}{9}\sum_{n=0}^\infty x^n \end{align*}$$ giving an answer of $$\frac{17}{72}\binom{50}{2} - \frac{52}{72}\binom{51}{2} + \frac{47}{72}\binom{52}{2} + \frac{1}{8} - \frac{1}{9} = 234$$ for the first problem.

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For the second problem, you would have $d,e,f > 0$, or by another substitution of $d' = d-1, e'=e-1, f'=f-1$ we'd be counting the number of triples $d',e',f' \geq 0$ such that $d' + 2e' + 3f' = 44$, so it's just the coefficient of $x^{44}$ of the same generating function. This gives $$\frac{17}{72}\binom{44}{2} - \frac{52}{72}\binom{45}{2} + \frac{47}{72}\binom{46}{2} + \frac{1}{8} - \frac{1}{9} = 184$$

Answer 3

We set the followings:

$$ \begin{align} a&=x+y+z\\ b&=x+y\\ c&=x\\ \\ 0&\leq x,y,z\\ \\ 50&=3x+2y+z \end{align} $$

First let’s evaluate the number of solutions with even $x$

$$ \begin{align} x&=2k\\ k&\in(0,1,…,8)\\ y&\in(0,…,25-3k) \end{align} $$

Next, let’s evaluate the number of solutions with odd $x$

$$ \begin{align} x&=2k+1\\ k&\in(0,…,7)\\ y&\in(0,…,23-3k) \end{align} $$

That bring our total solutions to

$$ \sum_{k=0}^{8}{\left(26-3k\right)}+\sum_{k=0}^{7}{\left(24-3k\right)}=234 $$

For the second case, simply solve for $3x+2y+z=44$. Hope you understand enough to work on it yourself