<s>[INST] How many positive integer solutions does the equation $a+b+c=100$ have if we require $aHow many positive integer solutions does the equation $a+b+c=100$ have if we require $a<b<c$?
I know how to solve the problem if it was just $a+b+c=100$ but the fact it has the restriction $a<b<c$ is throwing me off.
How would I solve this?
 [/INST] First we count the number of triples of positive integers $(a, b, c)$ with $a + b + c = 100$. There are $\binom{99}{2}$ of them, which results from an elementary application of stars and bars. Just line up a hundred dots, place a bar in between the $a$th and $(a +1)$th dot, and then place another bar $b$ dots to the right of the first one. Then $c$ is the number of dots to the right of the second bar. There are 99 spaces in between a hundred dots and you're choosing two of them to place bars in.
This is an overcount since it doesn't account for the condition $a < b < c$. First we should subtract the number of triples in which $a, b, c$ are not all distinct. There are 49 triples such that $a = b$: to count them, choose $x$ with $ 0 < x<50$, set $a = b = x$ and $c = 100-2x$. This accounts for all possible cases since if $a = b \ge 50$ then $c$ is not positive. There are also 49 triples with $a = c$, or with $b = c$, respectively. There are no triples with $a = b = c$, so the total number of triples with $a , b, c$ not all distinct is $3*49 = 147$.
The number of triples of positive integers $(a, b, c)$ with $a, b, c$ distinct and $a + b + c = 100$ is therefore $\binom{99}{2} -147$. Given a triple $(a, b, c)$ with $a< b < c$ and $a + b + c = 100$, there are six distinct ways to permute $a, b, c$, all of which our current estimate accounts for. Therefore we should divide by 6 to throw out all the triples that are in the wrong order. The final answer is $\frac{1}{6}(\binom{99}{2} -147) = 784$.
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