The Plugboard of an Enigma machine was by far the largest contribution to the number of keys for encrypting messages. If I am not mistaken, given an alphabet of \(n\) letters, the number of ways of connecting, thereby swapping, \(m\) pairs of letters can be computed using this formula:

\(\qquad\displaystyle{n \choose 2m}\prod_{0\le k< m}(2m - (2k+1))\)

Or, equivalently, \(\displaystyle{n \choose 2m}\frac{(2m)!}{2^{m}\cdot m!}\)

During the second world war, the initial Enigma machine used had a configuration of \(n=26\) and \(m=6\). So the number of ways of swapping pairs of letters was:

\(\qquad\displaystyle{26 \choose 12}\times 11\times 9\times 7\times 5\times 3\times 1 = 100,391,791,500\)

But why did Arthur Scherbius, the inventor of the Enigma machine, chose 6 as the number of pairs for the plugboard?

Suppose the plugboard could fit any number of cables (from 1 to 13), what is the relative strength of swapping 6 pairs of letters, compared to, say, 11 pairs? To find out, we can plugin in different values to the above formula. For example, using Sagemath:

n = 26
m = int(n/2)

for i in (1..m):
    print i,'\t', binomial(n, 2*i) * prod( (2*i-(2*j+1)) for j in (0..(i-1)) )

 Pairs   Ways of connecting
     1                  325
     2               44,850
     3            3,453,450
     4          164,038,875
     5        5,019,589,575
     6      100,391,791,500
     7    1,305,093,289,500
     8   10,767,019,638,375
     9   53,835,098,191,875
    10  150,738,274,937,250
    11  205,552,193,096,250
    12  102,776,096,548,125
    13    7,905,853,580,625

As we can see, had the Enigma machine been equipped with a plugboard with 11 cables instead of 6, it would have increased the size of the key space by over two thousand times. On the other hand, it's a waste of resources to have more than 11 cables :)