You are thinking along the right lines, but start by thinking about your statement regarding the first cord. There are 26 ways to choose the first letter, and 25 ways to choose the second letter, but the order of choice doesn't matter. This is the difference between a combination (number of ways of choosing where order doesn't matter) and a permutation (number of ways of choosing where order does matter). Choosing any 2 from 26 in which order doesn't matter gives 26X25/2 choices, not 26X25. For example, connecting D to Y is the same as connecting Y to D, so 26X25 is double counting. The same applies to all cords, so when you connect the next cord there will be 24X23/2 ways of connecting it. Ultimately, that will result in a factor of 2^N on the denominator if there are N cords.
But that's not all! Does it matter what order you put the cords into the plugboard? eg if I connect D to Y first, then I connect H to L next, would it change anything if I had connect H to L first, then D to Y? No, it wouldn't. In other words, the order in which you connect the plugboard leads doesn't matter. So, you need to further divide your answer by the number of different possible orders of connecting N leads. That is N factorial (written N!)
So, combining these two effects, your answer will be a factor of N!*2^N larger than Mr Sale's answer. For 10 leads, 2^10 = 1024, and 10! = 3628800 giving a factor of 3.7 billion (you mean 3.6 billion, not trillion).
So Mr Sale made no mistake, but the reasons he is correct are slightly subtle.
You are thinking along the right lines, but start by thinking about your statement regarding the first cord. There are 26 ways to choose the first letter, and 25 ways to choose the second letter, but the order of choice doesn't matter. This is the difference between a combination (number of ways of choosing where order doesn't matter) and a permutation (number of ways of choosing where order does matter). Choosing any 2 from 26 in which order doesn't matter gives 26X25/2 choices, not 26X25. For example, connecting D to Y is the same as connecting Y to D, so 26X25 is double counting. The same applies to all cords, so when you connect the next cord there will be 24X23/2 ways of connecting it. Ultimately, that will result in a factor of 2^N on the denominator if there are N cords.
But that's not all! Does it matter what order you put the cords into the plugboard? eg if I connect D to Y first, then I connect H to L next, would it change anything if I had connect H to L first, then D to Y? No, it wouldn't. In other words, the order in which you connect the plugboard leads doesn't matter. So, you need to further divide your answer by the number of different possible orders of connecting N leads. That is N factorial (written N!)
So, combining these two effects, your answer will be a factor of N!*2^N larger than Mr Sale's answer. For 10 leads, 2^10 = 1024, and 10! = 3628800 giving a factor of 3.7 billion (you mean 3.6 billion, not trillion).
So Mr Sale made no mistake, but the reasons he is correct are slightly subtle.