The possible combinations of $ (a,b,c)$ are $ a+b+c, c+ab, b+ca, a+bc,(b+c)a,(c+a)b,(a+b)c,abc$. If $ a<b<c$, then $ a+b+c < c+ab < b+ca < a+bc$, $ (b+c)a<(c+a)b<(a+b)c<abc$, $ b+ca<(c+a)b$, $ a+bc<(c+a)b$. So some combinations are equal iff $ a+bc = a(b+c)$. (a) If $ \frac{n}2<a<b<c\le n$, then $ a>\frac{c}2$. So $ a(b+c-1) > \frac{c}2(b+b) = bc$, i.e., $ a(b+c) > a+bc$. Thus all combinations are distinct. (b) Let $ p<b<c$ be the chosen numbers. Also let $ d$ be the positive integer such that $ b=p+d$. We have $ p+bc = p(b+c)$. Therefore $ p+ (p+d)c = p(p+c+d)$ and we get $ c = p+\frac{p(p-1)}d$. But $ b<c$, so $ p+d < p+\frac{p(p-1)}d$, i.e., $ d^2<p(p-1)$. Therefore $ d<p$. But $ d$ must divide $ p(p-1)$, so $ d$ divides $ p-1$. Clearly, for any divisor of $ p-1$, we get a satisfying triple. So we are done.

Босинг брат ман хам шундай ишладим

Listing out all of the possibilites with $a<b<c$, we see that the only way two combinations are equal is if $a(b+c)=a+bc$. (a) If $\tfrac{n}{2}<a<b<c<n$, we see that $a>\tfrac{c}{2}$. Thus, \[a(b+c)-a > \frac{c}{2} (b+c-1) \ge \frac{c}{2} \cdot 2b = bc.\] (b) Say the numbers are $p<q<r$, with $p+x = q$. We need \[p(p+x+r) = p+(p+x)r,\] which rearranges to \[r = \frac{p^2-p}{x}+p.\] Moreover, we know that $q<r$, so we can bound $x<p$. Thus, $x \mid p-1$, and each divisor of $p-1$ corresponds to a valid triple. $\square$