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If there are n elements in a set the number of binary operations that can be defined are 2n, am I right or wrong ?

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If the definition of "operation" is any function $\oplus \colon A \times A \to A$ without any further restrictions, there are $n^2$ values of the function a define independently among $n$ values, so it is $n^{n^2}$.

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    $\begingroup$ It can be also seen as the number of ways to fill an $n^2$-tuple where each position assumes $n$ values. $\underbrace{(\_\ \_\ \_ \ \_ \dots \_ \ \_)}_{n^2\ positions}$ So it is $n^{n^2}$ $\endgroup$ – nickchalkida Sep 6 '15 at 9:09
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Binary operations without any restriction? N0, there are many more. For every pair $(x,y)$ from a set $X$ we pick any value $f(x,y) \in X$, so we have $n^2$ many independent choices from $n$, so $n^{n^2}$ in my reckoning.

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