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My understanding of n-tuple and a permutation of n elements is, that both are ordered sequences of n elements.

Are there differences in the objects correlating to these two terms ?

I guess it would be nice if any help could be coded in language than can be understood by someone fascinated in and swamped by mathematics, though any help is appreciated.

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An $n$-tuple of elements in a set $X$ is a function $f:\{1,\ldots, n\}\rightarrow X$, whereas a permutation is a bijection $g:X\rightarrow X$. The differences are that an $n$-tuple can have any codomain and need not be a bijection (or even an injection or surjection), while a permutation has both the domain and codomain equal and is a bijection, so they are very different objects in general.

For example, we can represent an $n$-tuple as $(x_1,\ldots, x_n)$. But because the $n$-tuple is any function as above, we could have $x_i=x_j$ for $i\neq j$ (i.e. not injective) and there is always an issue of surjectivity, as we could take the $n$-tuple with every entry being equal; if $X$ has more than one element, this won't be surjective. On the other hand, a permutation of $\{1,\ldots, n\}$ is an $n$-tuple of elements in $\{1,\ldots, n\}$ such that we never have $x_i=x_j$ for $i\neq j$ (this ends up being sufficient for surjectivity as well).

However, in the case where we deal with permutations of finite sets, there is an instance in which they become similar. When $X$ is a finite set, we can restrict to the case where $X=\{1,\ldots,n\}$ for some natural number $n$: then an $n$-tuple of elements of $X=\{1,\ldots, n\}$ is a function $f:\{1,\ldots, n\}\rightarrow \{1,\ldots, n\}$. However, this still isn't necessarily a permutation, since we require that the function $f$ be a bijection. Thus, you could think of permutations as special cases of $n$-tuples on a finite set of size $n$.

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  • $\begingroup$ In the special of case your third paragraph: only if the n-tuple happens to be a bijection will it have the chance to generate the same sequence as a permutation of the same set ? Is this right ? $\endgroup$ – pidoretroma Jun 20 '14 at 17:38
  • $\begingroup$ Yes, only when it is a bijection does it qualify to be a permutation. $\endgroup$ – Hayden Jun 20 '14 at 17:41
  • $\begingroup$ @MJD Yes, you're correct. I meant "a permutation of $\{1,\ldots, n\}$ is..." Fixed $\endgroup$ – Hayden Jun 20 '14 at 17:43
  • $\begingroup$ Thanks a lot. My brain likes you more than wikipedia. $\endgroup$ – pidoretroma Jun 20 '14 at 17:45
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A n-tuple is a sequence $(x_1, \cdots, x_n)$. A permutation of elements of a set $A$ is a bijection $\sigma:A \rightarrow A$, and for finite sets, we can represent it by matrices. For example, if $A = \{1,2,3,4\}$, then $$\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 1 & 4 \end{pmatrix}$$ represents $\sigma(1) = 2, \sigma(2) = 3, \sigma(3) = 1, \sigma(4) = 4$. By heavily abusing the notation, we could think $\sigma(1,2,3,4) = (2,3,1,4)$, getting another n-tuple, but the objects are different. You can read more about this here.

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  • $\begingroup$ Your link tells me how you can define and present the permutation. but my question remains the same: if I have the 4-tuple (2,3,1,4) where is its difference to the specific permutation (2,3,1,4) of the set (1,2,3,4). The number of objects matters in both, so does their order. $\endgroup$ – pidoretroma Jun 20 '14 at 17:20

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