Would a comma have hurt anyone?

You are breaking into a vault that contains ancient Roman treasure. The vault is locked, and can be opened via a modern-day keypad. The keypad contains three numerical inputs, which are (of course) expressed using Roman numerals: “I,” “II,” and “III.”

It’s a good thing your accomplice was able to steal the numerical key code to the vault. Earlier in the day, they handed you this code on a scroll of paper. Once at the keypad, you remove the scroll from your pocket and unfurl it. It reads: “IIIIIIIIII.” That’s ten vertical marks, without any clear spacing between them.

With some quick mental arithmetic, you realize the combination to unlock the door could be anywhere from four digits long to 10 digits long. (Or is it IV digits to X digits?) How many distinct combinations are possible? If two combinations use the same numbers but in a different order, they are considered distinct.

Let $n_1$ be the number of "I" numerals, $n_2$ the number of "II" numerals and $n_3$ the number of "III" numerals. Since there are total of ten vertical marks, we must have $n_1 + 2n_2 + 3n_3 = 10.$ If we have a nonnegative solution to this Diophantine equation, then since any order of the different numbers is a distinct combination, then for any solution there are a total of $$\frac{(n_1+n_2+n_3)!}{n_1! n_2! n_3!}$$ possible combinations associated with the solution $(n_1, n_2, n_3).$

We can solve this Diophantine equation by iterating through the possible cases for $n_3 \in \{0,1,2,3\}.$ When $n_3 = 0,$ the remaining equation is $$n_1 + 2n_2 = 10,$$ which has nonnegative solutions $$n_1 = 10 - 2k, n_2 = k, k = 0, 1, 2, 3, 4, 5.$$ When $n_3 = 1,$ the remaining equation is $$n_1 + 2n_2 = 7,$$ which has nonnegative solutions $$n_1 = 7-2k, n_2 = k, k = 0, 1, 2, 3.$$ When $n_3 = 2,$ the remaining equation is $$n_1 + 2n_2 = 4,$$ which has nonnegative solutions $$n_1 = 4-2k, n_2 = k, k = 0, 1, 2.$$ Finally, when $n_3 = 3,$ the remaining equation is $n_1 + 2n_2 = 1,$ which has only a single nonnegative solution $n_1 = 1, n_2 = 0.$ So overall, the solution set is \begin{align*}\mathcal{N} &= \{ (0,2,2), (0,5,0), (1,0,3), (1,3,1), (2,1,2), (2,4,0), (3,2,1), \\ &\quad\quad (4,0,2), (4,3,0), (5,1,1), (6,2,0), (7,0,1), (8,1,0), (10,0,0) \}\end{align*} So, the total number of possible combinations is \begin{align*}N &= \sum_{(n_1,n_2,n_3) \in \mathcal{N}} \frac{(n_1+n_2+n_3)!}{n_1! n_2! n_3!} \\ &= \frac{4!}{0!2!2!} + \frac{5!}{0!5!0!} + \frac{4!}{1!0!3!} + \frac{5!}{1!3!1!} + \frac{5!}{2!1!2!} + \frac{6!}{3!2!1!} \\ & \quad\quad +\frac{6!}{4!0!2!} + \frac{7!}{4!3!0!} + \frac{7!}{5!1!1!} + \frac{8!}{6!2!0!} + \frac{8!}{7!0!1!} + \frac{9!}{8!1!0!} + \frac{10!}{10!0!0!} \\ &= 6+1+4+20+30+15+60+15+35+42+28+8+9+1 \\ &=274\end{align*}