imc

The cost of a rose is odd, hence we need an even number of roses. Let there be $2r$ roses for some $r\ge 0$. Then we have $50-3\cdot 2r=50-6r$ dollars left. We can always reach the sum exactly $50$ by buying $(50-6r)/2=25-3r$ carnations. Of course, the number of roses must be such that the number of carnations is non-negative. We get the inequality $25-3r\ge 0$ which solves to $r\le 8\frac{1}{3}$. $r$ must be an integer, so there are $\boxed{9}$ possible values of $r$, and each gives us one solution.