jmc

For any $n$, we note that $n!!$ is the product of a set of odd integers and is thus odd. For $n\ge 5$, then $n!!$ is divisible by $5$. Thus, the units digit of $n!!$ must be $5$ for $n\ge 5$. Hence, the units digit of the sum $5!!+7!!+\cdots +49!!$ is units digit of the sum of $5$, $\frac{49-5}{2}+1=23$ times. The units digit of $23\times 5$ is also $5$. Now, we have to sum this with $3!!+1!!=3\times 1+1=4$, yielding that the answer is $4+5=\boxed{9}$.