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This is small but quirky idea that popped into my head in the middle of the night last night.

If I have $n$ objects, and want to find out how many permutations (sequences) of $r$ objects there are, we use:

$_nP_r = \frac{n!}{(n-r)!}$.

Say for some reason, I want $r = 0$; that is, I want to see how many sequences I can make of length $0$.

$_nP_r = \frac{n!}{(n-r)!} = \frac{n!}{(n-0)!} = \frac{n!}{n!} = 1$

But shouldn't the answer be $\infty$? After all, couldn't you argue that there infinitely many sequences of length zero to be made from $n$ objects?

Obviously my thinking is flawed, but I'm interested in hearing from other perspectives why it is so.

Monica Heddneck
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    I don't see how you can say there are infinitely many sequences of 0 objects to be made from $n$ objects. There is only one sequence of 0 objects: the empty sequence. – Ted Apr 01 '16 at 06:13
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    Since you claim there are infinitely many, I would like you to list 10 please. – Em. Apr 01 '16 at 06:17
  • @Ted: good point...I guess that about wraps up my midnight meanderings. – Monica Heddneck Apr 01 '16 at 06:40

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