Enigma - How many possibilities does the plugboard have? I try to relate my own calculation with the real result being about 150.738.274.937.250.
The plugboard has 26 letters, and you have 10 wires to connect pairs of letters.
My idea was to calculate first how many possible pairs you can have.
26*25/2 = 325 [(a,a) not allowed, and (a,b)=(b,a) so divide by 2 to remove duplicates]
or just 26C2 = 325.
And out of 325 pairs to choose 10 pairs there are 325C10 = 3.150.234.554.696.452.080 possibilities to connect pairs with 10 wires.
Where is my way of thinking wrong?
out of 325 pairs to choose 10 pairs, there are 325C10 possibilities to connect pairs with 10 wires.

This accounts for the fact that we can not choose a pair that has both letters identical to an earlier pair, but misses that we can not choose a pair that has any letter common with an earlier pair.
A correct line of thinking is: imagine that the 20 ends of the 10 wires are numbered, and we assign a letter incrementally to each of them. That's 20 distinct letters to choose among 26, and order matters, giving $26!/(26-20)!$ ways. But the order of the 10 pairs does not matter, thus divide by $10!$. Further, the order for the two letters assigned to any of the 10 wires does not matter, thus divide by $2^{10}$. We end up with $26!/(6!\,10!\,2^{10})=150,738,274,937,250$ distinct ways.