Permalink Submitted by Gordon Elliott on August 17, 2018

Somewhere I put in a set of notes, and at the end did calculation. Viewing the coding sheet, clearly they used 10 cables with pairs of banana plugs at each end, and wires were swapped so that up to 10 pairs could be swapped. However they show entries in the coding sheet of "*" which I assume means no cable used--thus a jumper. In all places where a cable is not used, a jumper plug must have been used to map the output to itself. The "10" clearly comes from the coding sheet. I did the math wrong wherever my post appears.

This time, assume no "*" entries so all combinations from 10 cables are used. You don't count the remaining slots as combinations--they are algorithmically determined as jumper blocks to themselves so no choices! Since pairs swap and first choice is 26 starts to 25 remaining but swapping is equivalent, this means divide by 2.

## Pairs of letters in plugboard (Error correct from wherever)

Somewhere I put in a set of notes, and at the end did calculation. Viewing the coding sheet, clearly they used 10 cables with pairs of banana plugs at each end, and wires were swapped so that up to 10 pairs could be swapped. However they show entries in the coding sheet of "*" which I assume means no cable used--thus a jumper. In all places where a cable is not used, a jumper plug must have been used to map the output to itself. The "10" clearly comes from the coding sheet. I did the math wrong wherever my post appears.

This time, assume no "*" entries so all combinations from 10 cables are used. You don't count the remaining slots as combinations--they are algorithmically determined as jumper blocks to themselves so no choices! Since pairs swap and first choice is 26 starts to 25 remaining but swapping is equivalent, this means divide by 2.

Thus: 26*25 * 24*23 * ... * 10*9 * 8*7 / 2 = 26!/(6!*2) = 280.0635 * 10^21 ish.