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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?

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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.

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The above answer is correct, but I know from teaching this problem to others some still have trouble grasping why it is right. If that is you, think of it this way:

You have ten wires. Consider them each in turn.

For the first wire, you have 26 choices of where to plug in one of its ends. After doing so, you have 25 choices where to plug the other end. Two holes have now been filled. Therefore, there are 26 * 25 ways to plug in the first wire.

For the second wire, you have 24 choices left where to plug in one of its end. After doing so, you have 23 choices where to plug in the other end. Four holes have now been filled in total. Therefore, there are 26 * 25 * 24 * 23 ways to plug in the first TWO wires. Are you following so far?

If you continue this approach with the 3rd, 4th, 5th, 6th, 7th, 8th, 9th, and 10th wire, you will determine that there are (26 * 25) * (24 * 23) * (22 * 21) * (20 * 19) * (18 * 17) * (16 * 15) * (14 * 13) * (12 * 11) * (10 * 9) * (8 * 7) ways to plug in the ten wires. Note there are ten groups of parentheses. They are unnecessary and only shown for pedagogical purposes as each group of parentheses represents one of the ten wires you plugged into the plugboard. Hopefully, this makes sense to you so far. What is this number? It is the exact same as 26!/6!, which is the part of the expression in the answer above. I am trying to more fully explain where 26!/6! came from, as this is the part that seems to confuse some people who are new to combinatoric problems.

Next, as the answer above mentions, you need to consider the order that you plugged in the wires doesn't matter. Let's say you chose to plug the first wire into letters 'A' and 'B' and the second one into 'C' and 'D'. It doesn't matter if you plugged wire one into letters 'A' and 'B' first and then wire two into 'C' and 'D' or if you instead plugged wire two into 'C' and 'D' first and then plugged wire one into 'A' and 'B'. The result would be identical, so the order you plug the ten wires into the plugboard doesn't matter. Consequently, you have to divide 26!/6! by the total number of ways you could have ordered the plugging in of the ten wires. Maybe you plugged in wire one first, then wire five, then wire ten, etc. But, again, the order you chose to use for plugging in the ten wires doesn't matter because the ten wires are identical to each other. Since there are ten wires, there are 10! ways you could have ordered them. Because this ordering doesn't matter, you must now divide 26!/6! by 10! to eliminate all of the inconsequential ordering of how the ten wires were plugged in. And now we're almost to the correct answer. The next step is you now must consider whether it matters which end of the wire you plugged into a particular plugboard hole. It could matter if, for instance, the wire contained a diode that only allows current to flow in one direction. But we are using simple wires so the ends are functionally identical. Therefore, we must divide our expression by 2 for each wire we used because it doesn't matter if we wired 'A' to 'B' or 'B' to 'A', the only thing that matters is that two particular letters are connected by a wire. So we must divide the expression by 2 a total of ten times, once for each of the ten wires. Therefore, the expression is now 26!/(6! * 10! * 2^10), which is the answer given above.

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  • $\begingroup$ Welcome to crypto-SE. Notice that in your answer, "above answer" is uncertain: should your answer become the most popular, it will rise above. And when there is a tie and no answer is accepted, the order is random. The best solution to this issue is a link to the answer, as obtained using the "share" button on the answer. A simpler option is to say "the accepted answer", though this can (more rarely) change too. $\endgroup$
    – fgrieu
    Mar 12 at 9:52

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