Say $f(n)$ is number of bijections in a set of $n$ elements without a fixed point. What would this mean? I know that a bijection means that each element in one set is paired with exactly one element in another set. But what does without a fixed point mean?
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3As an aside, a bijection with no fixed points is referred to as a Derangement. The number of derangements on a set with $n$ elements is often denoted as $!n$ or $D_n$ or something similar. – JMoravitz Oct 16 '16 at 21:15
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Thanks for mentioning this piece of knowledge! – PiccolMan Oct 16 '16 at 21:23
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1It means f(x) is never equal to x. – fleablood Oct 17 '16 at 06:56
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A fixed point of a function $f:X\to X$ is an element $x\in X$ such that $f(x)=x$.
For instance, if $X=\{1,2,3\}$, then the function defined by $f(1)=1,f(2)=3,f(3)=2$ has $1$ as a fixed point, while the function $g(1)=2,g(2)=3,g(3)=1$ has no fixed points.
carmichael561
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A fixed point is of the form $f(n) = n$. It means that $f$ fixes the "location" of $n$. If there are no fixed points, then every element in the set is "moved".
MathMajor
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Yet another point of view.
Consider the classical association of a permutation on $n$ objects with its matrix, for example:
$$\text{Permutation} \ \ \binom{1 \ 2 \ 3}{2 \ 3 \ 1} \ \ \ \text{is associated with} \ \ \ \pmatrix{0&0&1\\1&0&0\\0&1&0}$$
Thus, clearly, a permutation without any fixed point (like the one above) is associated with permutation matrices having only zeros on their diagonal.
Jean Marie
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A permutation of a set of $n$ elements that has no fixed point is called a derangement. In this answer, it is shown that the number of derangements of a set of $n$ elements is $$ \mathcal{D}_n=n!\sum_{k=0}^n\frac{(-1)^k}{k!} $$ which is the closest integer to $\dfrac{n!}e$ for $n\ge1$.
