The Enigma Machine

Solution:

This problem is mix of combinatorics and programming, but some basic knowledge from both are ultimately necessary for successful solving, thus I will leave few links that can be helpful and that will back up my solution.

Links: combinations , permutations and probably this, permutations with restriction .

1. So let's begin. Let's begin with Rotor placement (step choosing rotors and placing rotors combined), as it can be seen, there are $X$ objects (rotors), which will be somehow arranged in $R$ designated positions, with guarantee $X \ge R$ - these are permutations with restriction (if $X > R$ ) or permutations (if $X = R$ ), in both case, generalization can be made and it can be said that it can be calculated with the following: $\left( \begin{matrix} X \\ X-R \end{matrix} \right)$ .

2. Initial rotation of each rotor (or initial setting of each rotor ): I believe it is obvious that each rotor, having $K$ letters, can be set to $K$ different positions, and having $R$ rotors placed into the machine, that is: $K^{R}$

3. Finally the plugboard (since 4. will heavily rely on 3). There are $K = 2 \times w$ letters, since $K$ is even. As the text of problem states, practically pairs need to be formed, but the trick is, that number of pairs varies as well. Let's introduce $j \in \{0, 1, ... w\}$ , which represents number of pairs, and this iterator will take each value from the set. Forming first pair is done on $\left( \begin{matrix} K \\ 2 \end{matrix} \right)$ ways, forming second pair $\left( \begin{matrix} K-2 \\ 2 \end{matrix} \right)$ , and so on. Since the order of pairing is not important (if we paired letters ab and cd , it is completely irrelevant whether we firstly formed pair ab or cd ), division with $j!$ is needed. Putting all that in one formula, result is: $\sum _{j=0} ^{\frac{K}{2}} { ( \frac{K!}{2^{j} \times j! \times (K - 2\times j)!} ) }$ .

4. Reflector setting - this is basically sub-case from 3. where $j = \frac{K}{2}$ . Thus, solving this is easier than whole part 3. and the formula is: $\frac{K!}{2^{j} \times j! }$ .

The results of each segment are multiplied, and that is the final result. Also, if variable $P = 0$ , solution for part 3 is $1$ instead, the same goes for variable $A$ if zero, then solution for part 4 is $1$ . If everything is carefully converted into programming code, the program should generate $\boxed{343038620}$ as an answer for the given file.