To see how this is worked out you must know some basic facts about permutations and combinations.
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.)
n! / r! (n-r)!
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:
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
Again, if you don't believe this can happen, have a look at the number of ways you can put 2 pieces of wire into 6 plugboard sockets: there are 45 ways, three times as many as you get with 3 wires. (Hint: imagine pulling out one of the three wires.)
Note contributed by Andrew Hodges (based on page 178 of his book Alan Turing: the Enigma.