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What is the number of one-to-one functions $f$ from the set $\{1, 2, \ldots, n\}$ to the set $\{1, 2, \ldots, 2n − 1\}$ so that $f(x) \neq 2x − 1$ for all $x$?

I'm not sure if I did the question correctly, so please provide feedback. The total number of one-to-one functions is $(2n-1)(2n-2) \ldots (n)$ or $\frac{(2n-1)!}{(n-1)!}$. The number of one-to-one functions such that $f(x) = 2x-1$ is $n$. Thus, the total number of one-to-one functions such that $f(x) \neq 2x-1$ is: $$\frac{(2n-1)!}{(n-1)!} - n$$

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    $\begingroup$ You want to discard the number of one to one functions such that $f(x)=2x-1$ for some $x$. Those are way more than $n$, for instance just considering the number of one to one functions such that $f(1)=1$ this is $(2n-1)!/(n-1)!$ $\endgroup$ – Julian Mejia Apr 19 at 2:42
  • $\begingroup$ How would you go about finding the number of one-to-one functions such that f(x) = 2x-1? $\endgroup$ – Brownie Apr 20 at 22:08

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