Would a comma have hurt anyone? Extra Credit

Having successfully hacked your way through the first keypad, the door opens to reveal a second door with yet another keypad that has six numerical inputs: “I,” “II,” “III,” “IV,” “V,” “VI,” “VII,” and “VIII.”

You were expecting this, which is why your accomplice had handed you a second scroll of paper. You unfurl this one as well, hoping they remembered to add spaces between the numbers.

No such luck. This paper reads: “IIIVIIIVIIIVIII.” That’s 15 characters in total. How many distinct combinations are possible for this second door?

Since there are three V characters, let's iterate on the various possibilities for these characters. Let's write the code word as $$C = R(w_1)R(x_1)R(w_2)R(x_2)R(w_3)R(x_3)R(w_4),$$ where $R(t)$ be the Roman numeral of the number $t,$ where we see that $w_i \in \{1, 2, 3\}$ and $x_i \in \{ 4, 5, 6, 7, 8 \}.$ In order for the our word $C$ to match the code provided, we would need to make sure that if $x_i = 4$ then $x_{i-1} \leq 7,$ for $i = 2, 3.$ Let $$\mathcal{X} = \{ (x_1, x_2, x_3) \in \{4, 5, 6, 7, 8 \}^3 \mid x_i = 4 \Rightarrow x_{i-1} \leq 7, i = 2, 3 \}$$ be the solution set for the values of the V characters.

Let's define the indicator function $$\delta(t) = \begin{cases} 1, &\text{if $t=1$}\\ 0, &\text{if $t=0$.}\end{cases}$$ Additionally let $N(t)$ be the number of possible combinations of numerals that in "I", "II", and "III" that can be found in $R(t).$ In particular, $N(0)=N(1)=1,$ $N(2)=2$ and $N(3)=4.$ For any $(x_1, x_2, x_3) \in \mathcal{X},$ we would also need to have the following conditions be met in order for our word $C$ to match the code provided: \begin{align*}w_1 = w_1(x_1) &= 3 - \delta(x_1-4) \\ w_2 = w_2(x_1,x_2) &= \min \{ 3, 8 - x_1 \} - \delta(x_2 - 4) \\ w_3 = w_3(x_2,x_3) &= \min \{ 3, 8 - x_2 \} - \delta (x_3-4) \\ w_4 = w_4(x_3) &= \min \{3, 8 - x_3 \}\end{align*} Then, the total number of distinct code combinations where the V characters are assigned to the numbers $(x_1,x_2,x_3)$ is given by $N(w_1(x_1))N(w_2(x_1,x_2))N(w_3(x_2,x_3))N(w_4(x_3)).$

Therefore, the total possible number of distinct combinations for the second door is $$\hat{N} = \sum_{(x_1,x_2,x_3) \in \mathcal{X}} N(w_1(x_1))N(w_2(x_1,x_2))N(w_3(x_2,x_3))N(w_4(x_3)) = 4000.$$