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Puzzle for Today

Puzzle No. 485 – Wednesday 22 May

A recent Today Programme puzzle found that with a set of 5 Enigma rotors there are a total of 5*4*3=60 different ways to pick a subset of 3 (as the final order matters). Enigma operation rules stated that using the same rotor in the same position two days in a row was not allowed. Assuming that Bletchley Park’s Codebreakers know which set of three rotors was used yesterday, how many can be discounted from the sixty options for today?

Today’s #PuzzleForToday has been set by Tom Briggs, education manager at Bletchley Park

Let’s say that yesterday’s rotor set was was I, II, III.

Anything that has I in the leftmost position can be discounted (there are 4*3 = 12 of those);

Anything that has II in the middle can be discounted (there are 4*3 = 12 of those as well, but 3 have already been counted as they also start with I, so we can only discount another 9).

Anything that has III in the rightmost position can be discounted (there are 4*3 of those too, but 3 have already been counted as they have a I on the left and two more have already been counted as they also have a II in the middle, so we can only discount another 7).

12 + (12 – 3) + (12 – 5) = 28 of the rotor orders can be immediately discounted.

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