Bertie

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UsernameBertie
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 #5
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It's a truly brilliant formula, I'd quite happily buy a drink for the person that constructed it. Like Nauseated I've tried to find out where it comes from but without success.

 

Back to the question, (the original one with 19 computers and 23 peripherals). It's badly worded. It's true that each computer should be connected to a peripheral and that each peripheral should be connected to a computer, but there's more to it than that. What it should also say is that any of the computers can be connected to several peripherals if need be, but each peripheral can be connected to only one computer.

 

An immediate consequence of that is if there are more computers than peripherals then no connection setup is possible, we run out of peripherals. Miraculously, in these cases, the formula produces a zero result. (I have a proof of that, I don't have the time to post it at the moment, but I shall do it tomorrow night or possibly Wednesday morning (UK), assuming that no-one else does so in the meantime). That also answers one of Melody's questions, if N>k then the result is zero.

 

If N=k the formula produces the expected result N!

 

If N<k, there will be an excess of peripherals over computers and in this case one or more of the computers have to be connected to more than one peripheral. The formula produces the number of (the most economical) possible setups

Consider a simple case of 3 computers and 5 peripherals, (before dealing with the 19,23 problem). The possible connection setups are (1) one of the computers is connected to three of the peripherals and the other two computers connected to one peripheral each, and (2) two of the computers are connected to two peripherals each and the third computer to the remaining peripheral. Do the arithmetic and you find that there are 150 possible connection setups and that this is also the result produced by the formula.

 

Have to go now.

Mar 30, 2015