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