Exercise 3: Counting Rotor Choice-and-Order Combinations — Possible Solution
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This is a permutations problem, not a combinations problem — ORDER
matters, because installing rotor II-III-I produces a completely
different machine from I-II-III, even though it's the same three
rotors.

Choosing and arranging 3 rotors out of 5 available, where order
matters:

    Position 1 (leftmost):   5 choices (any of the 5 rotors)
    Position 2 (middle):     4 choices (whichever 4 remain)
    Position 3 (rightmost):  3 choices (whichever 3 remain)

Total combinations = 5 x 4 x 3 = 60

So there are 60 distinct rotor choice-and-order combinations possible
from a set of 5 rotors.

WHY THIS WORKS AS AN ANSWER
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This is the standard "permutations of n taken k at a time" calculation
(P(5,3) = 5!/(5-3)! = 5x4x3 = 60), applied directly to the Walzenlage
setting described in the chapter's own Daily Key Settings section.
60 is only ONE of the four factors that combine into the chapter's
full ~10^114 keyspace figure — ring settings, starting positions, and
the roughly 150 trillion possible plugboard pairings each multiply
this 60 further, which is why the full keyspace is so many orders of
magnitude larger than this single factor alone.