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I want to write a formula for $n!$.

$n!$ is the number of permutation functions on the set $\{1,\ldots,n\}$.

Let's define a "true k-permutation" on $\{1,\ldots,n\}$ as a permutation that is identity on exactly $n-k$ elements, i.e., if $\phi$ is a $k$-permutation then $\phi(i) = i$ only on $n-k$ elements.

This seems like an elementary object yet I don't know what you call such perms (That is my question).

Notice that the relation $\phi$ is a true $k$-permutation partitions the group of permutations on $\{1,\ldots,n\}$.

So there is a formula for $n!$ involving counting of true $k$-perms. The one I have involves sum partitions of $k$ and disjoint compositions of cycles. But there's probably other ways to count them.

But this has to have been done already, so I would appreciate it if someone could post a link.

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It does not hurt if you had to type in permutation instead of perm. Reading that several times makes me feel weird! – user21436 Mar 14 '12 at 8:15
You should check the following link: – Dennis Gulko Mar 14 '12 at 8:19
And perhaps more relevantly this one:, which is linked to from the one Dennis mentioned. – Tara B Mar 14 '12 at 10:30
Your first sentence is puzzling. Surely you know a formula for $n!$ that doesn't require counting the permutations with $n-k$ fixed points (which is the usual way to designate your "true $k$ permutations") separately? – Marc van Leeuwen Mar 14 '12 at 10:41
Thanks @Arturo Magidin. I will typeset fully in future. – Enjoys Math Mar 14 '12 at 23:20
up vote 5 down vote accepted

A permutation which leaves no element fixed (what you call a "true $n$-permutation") is called a derangement. I know of no common term for "true $k$-permutations" in general.

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Well, probably they are usually just called permutations with support size $k$. (The support of a permutation $\sigma$ is the set of points moved by $\sigma$.) – Tara B Mar 14 '12 at 10:47

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