smc

We can factor the second equation to get $c(a+b)=23$, so we see that $c$ must be a factor of $23$, and since this is prime, $c=1$ or $c=23$. However, if $c=23$, then $a+b=1$, which is impossible for the field of positive integers. Therefore, $c=1$ for all possible solutions. Substituting this into the original equations gives $ab+b=44$ and $a+b=23$. From the second equation, $a=23-b$, and substituting this into the first equation yields $b(23-b)+b=44$, or $b^2-24b+44=0$. Factoring this gives $(b-2)(b-22)=0$, so $b=2$ or $b=22$. Both of these yield integer solutions for $a$, giving $a=21$ or $a=1$, respectively. Therefore, the only solutions are $(21,2,1)$ and $(1,22,1)$, yielding $2$ solutions, $\boxed{2}$.