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Research Note Description:	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. <br> For each of these the plugboard (with ten pairs of letters connected) can be in 150,738,274,937,250 possible states.<br> <b><font color=red>Counting the Possible Plugboard Settings</b></font><br> The exact figure for settings of the plugboard with 10 pairs of letters connected is <font color=blue>150,738,274,937,250. </font><br> To see how this is worked out,basic facts about permutations and combinations are given below: <br> • 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.) <br> • 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)! <br> 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. <br> 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.<br> C(26,2) x C(24,2) x C(22,2) x ... x C(8,2) <br> This can be simplified, with many factors cancelling, to <br> 26! / (6! 210) <br> 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:<br> 26! / (6! 10! 210) = 150,738,274,937,250.<br> More abstractly: the number of ways of choosing m pairs out of n objects is: <br> n! /((n-2m)! m! 2m) <br> If you want to convince yourself of this formula you might like to check that there are: <br> <ul><li>3 different ways of putting 2 pairs of wire into 4 plugboard sockets.</li></ul> <ul><li>15 different ways of putting 3 pairs of wire into 6 plugboard sockets.</li></ul> <br> 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: <br> 1 pair: 325<br> 2 pairs: 44.850<br> 3 pairs: 3,453,450<br> 4 pairs: 164,038,875<br> 5 pairs: 5,019,589,575<br> 6 pairs: 100,391,791,500<br> 7 pairs: 1,305,093,289,500<br> 8 pairs: 10,767.019,638,375<br> 9 pairs: 58,835.098,191,875<br> 10 pairs: 150,738,274,937,250<br> 11 pairs: 205,552,193,096,250<br> 12 pairs: 102,776,096,548,125<br> 13 pairs: 7,905,853,580,625<br><font color=gray>The total number of combinations is thus (even for the simplest military Enigma) of the order of 15,000,000,000,000,000,000</font> <p><center><img SRC="seminar_43.gif"></center> <p><b>Reference: </b> <a href= "http://www.codesandciphers.org.uk/enigma/enigma3.htm" target="_blank">www.codesandciphers.org.uk</a>
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Prof. Ashay Dharwadker