smc

The three operations can be performed on any of $3!=6$ orders. However, if the addition is performed either first or last, then multiplying in either order produces the same result. So at most four distinct values can be obtained. It is easy to check that the values of the four expressions $$\begin{array}{rl}(2\times 3)+(4\times 5)& =26,\\ (2\times 3+4)\times 5& =50,\\ 2\times (3+4\times 5)& =46,\\ 2\times (3+4)\times 5& =70\end{array}$$ are in fact all distinct. So the answer is $\boxed{4}$, which is choice $(\mathrm{C})$.