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The basic 3-rotor Enigma has 26x26x26 = 17,576 possible rotor states for each of 6 wheel orders giving 6x17,576 = 105,456 machine states.
For each of these the plugboard (with ten pairs of letters connected) can be in 150,738,274,937,250 possible states.
Counting the Possible Plugboard Settings
The exact figure for settings of the plugboard with 10 pairs of letters connected is
To see how this is worked out,basic facts about permutations and combinations are given below:
• Given n distinct objects there are n! ways of arranging them in sequence, where n! means the product n x (n-1) x (n-2)... 3 x 2 x 1. For example the six digits 1,2,3,4,5,6 can be arranged in 6 x 5 x 4 x 3 x 2 x 1=720 different orders (bell-ringers will be familiar with this.)
• Given a set of n distinct objects there are C(n,r) ways of dividing it into two sets of size r and (n-r), where C(n,r) means n! / r! (n-r)!
In the Enigma plugboard problem, the 26 letters have to be divided into 6 unpaired letters and 10 pairs of pairwise connected letters. One way of doing this is as follows: suppose that we had ten differently coloured connecting wires: red, blue, green etc etc.
Then there are C(26,2) ways of choosing a pair for the red wire. For each of these there are C(24,2) ways of choosing a pair for the blue wire, and so on, giving the product.
C(26,2) x C(24,2) x C(22,2) x ... x C(8,2)
This can be simplified, with many factors cancelling, to
26! / (6! 210)
But in the actual Enigma the wires are not coloured. This means we must divide by the number of ways of permuting the 10 coloured wires, i.e. divide by a further factor of 10!. This gives the answer:
26! / (6! 10! 210) = 150,738,274,937,250.
More abstractly: the number of ways of choosing m pairs out of n objects is:
n! /((n-2m)! m! 2m)
If you want to convince yourself of this formula you might like to check that there are:
3 different ways of putting 2 pairs of wire into 4 plugboard sockets.
15 different ways of putting 3 pairs of wire into 6 plugboard sockets.
From this formula we can find out something which often surprises people, which is that the number of possible plugboard pairings is greatest for 11 pairs, and then decreases:
1 pair: 325
2 pairs: 44.850
3 pairs: 3,453,450
4 pairs: 164,038,875
5 pairs: 5,019,589,575
6 pairs: 100,391,791,500
7 pairs: 1,305,093,289,500
8 pairs: 10,767.019,638,375
9 pairs: 58,835.098,191,875
10 pairs: 150,738,274,937,250
11 pairs: 205,552,193,096,250
12 pairs: 102,776,096,548,125
13 pairs: 7,905,853,580,625
The total number of combinations is thus (even for the simplest military Enigma) of the order of 15,000,000,000,000,000,000
Prof. Ashay Dharwadker