Question

just one thought about 0!=1.

Answer 1

I have been offered that `an empty set can be rearranged in 1 way`, and 0! (which means/is "the empty set") is therefore equal to 1.

I didn't like that, because it it (seems to me, as if it) is like saying that 0/1=1, or something along these lines. (maybe it is a very good idea)

the math I came up with is based on the premise that (for the least part) permutations and combinations are equivalent when choosing 0 things out of n things.

\(\LARGE\color{black}{{\rm \color{green}{_n}C\color{darkgoldenrod}{_0}}={\rm \color{green}{_n}P\color{darkgoldenrod}{_0}} }\)

\(\LARGE\color{black}{
\frac{\LARGE {\rm \color{green}{n}}! }{\LARGE{\rm (\color{darkgoldenrod}{0}!)\times(\color{green}{n}-\color{darkgoldenrod}{0})}! }=\frac{\LARGE {\rm \color{green}{n}}! }{\LARGE{\rm (\color{green}{n}-\color{darkgoldenrod}{0})}! } }\)

\(\LARGE\color{black}{
\color{darkgoldenrod}{0}!\times\frac{\LARGE {\rm \color{green}{n}}! }{\LARGE {\rm \color{green}{n}}! }=\frac{\LARGE {\rm \color{green}{n}}! }{\LARGE{\rm \color{green}{n}}! } }\)

\(\LARGE\color{black}{
\color{darkgoldenrod}{0}!=1 }\)

Answer 2

but I guess that (therefore) the idea that an empty set can be rearranged in one way is correct, or at least that is pretty much the same as the premise I used.