n/(n-1)/(n-2)/…/1 = n/(n-1)! = n*n/(n*(n-1)!) = n^2/n!
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Or just n/(n - 1)!
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Ramajuan product..cross multiply the factorial would be a neat idea..(n/n-(1/n)
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Thanks. Twitter will use this to make your timeline better. UndoUndo
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so obvious
Thanks. Twitter will use this to make your timeline better. UndoUndo
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Really? Tell me why 1/(n-1)/(n-2)/(n-3)/.../(1) = n/n!? And, why (n-1)/(n-2)/(n-3)/.../(1) = n!/n? It's impossible!
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Your error is that you have misinterpreted the divisions. 1/a/b = 1/(ab), not 1/(a/b).
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If you hardcode, in Assembler, you can use tricks like this to speed things up. I.E. on some CPUs 10*0.5 is a magnitude faster in terms of CPU cycles than 10/2
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Or just shift the binary to right
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'n' in the numerator and everything after a '/' goes in the denominator. It's the same as: n/((n-1)(n-2)...3.2.1)
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