The number of triples $(a, b, c)$ of [positive integers](https://artofproblemsolving.com/wiki/index.php/Natural_numbers) which satisfy the simultaneous [equations](https://artofproblemsolving.com/wiki/index.php/Equation)
$ab+bc=44$
$ac+bc=23$
is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ } 4$
We can [factor](https://artofproblemsolving.com/wiki/index.php/Factoring) the second equation to get $c(a+b)=23$, so we see that $c$ must be a [factor](https://artofproblemsolving.com/wiki/index.php/Factor) of $23$, and since this is [prime](https://artofproblemsolving.com/wiki/index.php/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, $\boxed{\text{C}}$.