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My question is how many permutations of {1,2,....n} have a single cycle?

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    What do you mean single cycle ? – 2017-01-23
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    There are two interpretations, the permutation is an $n$-cycle, or the permutation is any cycle of length $n$ or less. which one do you want? – 2017-01-23
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    @ openspace I don't know. – 2017-01-23
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    @ Jorge Fernandez Hidalgo the answer is n-1, but I don't know how? – 2017-01-23
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    isn't it $(n-1)!$? that makes way more sense – 2017-01-23
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    @ Jorge Fernandez Hidalgo I am new to this, would you give me some explaination? – 2017-01-23
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    You may want to read this part of this wikipedia page: https://en.wikipedia.org/wiki/Permutation#Cycle_notation – 2017-01-23
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    The cycle $(1,2,...,n)$ can be represented by $n-1$ transpositions, eg: $(1,2)(2,3)(3,4)...(n-1,n)$ or $(1,2)(1,3)(1,4)...(1,n)$. Maybe it is what you are looking for. – 2017-01-23

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A permutation is simply a reordering of a set into a different order - a bijection from the set to itself.

Every permutation has a decomposition into cycles. If it is a permutation of $1,2 \dots n$ then you get the decomposition by seeing that the permutation takes $1$ to $a$, and $a$ to $b$ and ... and $r$ to $1$. Then suppose that $4$ is the first integer not already named, $4$ goes to $s$, $s$ goes to $t$ ... and $y$ goes to $4$. And so on, until all $n$ integers have been named. Perhaps $z$ is a fixed point of the permutation. You end up with something like $(1,a,b,\dots r)(4, s, t \dots y)\dots (z)$, where each bracket contains one of the chains (often written without commas). Each bracket is called a cycle.

If you have just one cycle, you start with $1$, and go through all the $n-1$ other members of the set* before returning to $1$ again. Those other members can be arranged in $(n-1)!$ ways.

You should be able to show that there are in total $n!$ permutations.

Note: Because you are permuting a finite set and a permutation is one to one and onto (injective and surjective) it is easy to prove you will always get back to where you started. You must repeat eventually, and you can work backwards to a unique predecessor as well as forwards to a unique successor.

*(without [hesitation], repetition or deviation) - a comment which will be better understood in England than some other places

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    @ Mark Bennet In another word, because there are n choices for the first member in a cycle, then the rest has (n-1)!. Since the first is fix, it is not counted? but there are (n-1)! permutation all beginning with the same first member for a specific cycle? – 2017-01-23