Probability that a random permutation has no fixed point among the first $k$ elements

Is it true that $\frac1{n!} \int_0^\infty x^{n-k} (x-1)^k e^{-x}\,dx \approx e^{-k/n}$ when $k$ and $n$ are large integers with $k \le n$?

This quantity is the probability that a random permutation of $n$ elements does not fix any of the first $k$ elements.

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Here’s a heuristic idea, based on the strong law of large numbers (SLLN). A rigorous proof can be made using the central limit theorem (CLT), as in my new answer.

First write
$$I = \frac{1}{{n!}}\int_0^\infty {x^{n – k} (x – 1)^k e^{ – x} \,{\rm d}x} = \int_0^\infty {\bigg(1 – \frac{1}{x}} \bigg)^k \frac{{x^n e^{ – x} }}{{\Gamma (n + 1)}}\,{\rm d}x.$$
Now, if $X_1,\ldots,X_{n+1}$ are i.i.d. exponential(1) random variables, then their sum $X_1 + \cdots + X_{n+1}$ has gamma density $x^n e^{-x} / \Gamma(n+1)$, $x > 0$. Hence,
$$I = {\rm E} \bigg[ 1 – \frac{1}{{X_1 + \cdots + X_{n + 1} }}\bigg]^k = {\rm E}\bigg[1 – \frac{{n + 1}}{{X_1 + \cdots + X_{n + 1} }}\frac{1}{{n + 1}}\bigg]^k .$$
By SLLN, $(n + 1)^{ – 1} \sum\nolimits_{i = 1}^{n + 1} {X_i }$ converges almost surely to ${\rm E}[X_1 ] = 1$ as $n \to \infty$. So this suggests the approximation
$$I \approx \bigg(1 – \frac{1}{{n + 1}}\bigg)^k = \bigg(1 – \frac{1}{{n + 1}}\bigg)^{n(k/n)} \approx e^{ – k/n}.$$

A rigorous proof can be made using the central limit theorem (CLT), as follows.

Since
$$\frac{1}{{n!}}\int_0^1 {x^{n – k} (x – 1)^k e^{ – x} \,{\rm d}x} \to 0$$
as $n \to \infty$, it suffices to consider the integral from $1$ to $\infty$. Let $c>0$ be arbitrary but fixed, and let $X_i$ be as in my previous answer. Define $f_n {(x)} = x^n e^{-x} / \Gamma(n+1)$, $x>0$, and put $n'=n+1$. Then, since $f_n$ is the density of $\sum\nolimits_{i = 1}^{n'} {X_i }$,
$$\frac{1}{{n!}}\int_1^{n' – c\sqrt {n'} } {x^{n – k} (x – 1)^k e^{ – x} \,{\rm d}x} \leq \int_0^{n' – c\sqrt {n'} } {f_n (x)\,{\rm d}x} = {\rm P}\bigg(\frac{{\sum\nolimits_{i = 1}^{n'} {X_i } – n'}}{{\sqrt {n'} }} \le – c \bigg).$$
By the CLT, since the $X_i$ have unit mean and unit variance, the probability on the right-hand side tends to $\Phi(-c)$ as $n \to \infty$, where $\Phi$ denotes the distribution function of the standard normal distribution.
Similarly,
$$\frac{1}{{n!}}\int_{n' + c\sqrt {n'} }^\infty {x^{n – k} (x – 1)^k e^{ – x} \,{\rm d}x} \leq \int_{{n'} + c\sqrt {n'} }^\infty {f_n (x)\,{\rm d}x} = {\rm P}\bigg(\frac{{\sum\nolimits_{i = 1}^{n'} {X_i } – n'}}{{\sqrt {n'} }} > c \bigg),$$
and, by the CLT, the probability on the right-hand side tends to $1 – \Phi(c)$ as $n \to \infty$.
Next, note that
$$\bigg(1 – \frac{1}{{n' – c\sqrt {n'} }}\bigg)^k \int_{n' – c\sqrt {n'} }^{n' + c\sqrt {n'} } {f_n (x)\,{\rm d}x} \leq \frac{1}{{n!}}\int_{n' – c\sqrt {n'} }^{n' + c\sqrt {n'} } {x^{n – k} (x – 1)^k e^{ – x} \,{\rm d}x}$$
and
$$\frac{1}{{n!}}\int_{n' – c\sqrt {n'} }^{n' + c\sqrt {n'} } {x^{n – k} (x – 1)^k e^{ – x} \,{\rm d}x} \leq \bigg(1 – \frac{1}{{n' + c\sqrt {n'} }}\bigg)^k \int_{n' – c\sqrt {n'} }^{n' + c\sqrt {n'} } {f_n (x)\,{\rm d}x}.$$
By the CLT,
$$\int_{n' – c\sqrt {n'} }^{n' + c\sqrt {n'} } {f_n (x)\,{\rm d}x} = {\rm P}\bigg(-c \leq \frac{{\sum\nolimits_{i = 1}^{n'} {X_i } – n'}}{{\sqrt {n'} }} \leq c \bigg) \to 2 \Phi(c) -1$$
as $n \to \infty$. Combining it all, and noting that
$$\bigg(1 – \frac{1}{{n' \pm c\sqrt {n'} }}\bigg)^k = \bigg(1 – \frac{1}{{n' \pm c\sqrt {n'} }}\bigg)^{n(k/n)} \approx e^{ – k/n},$$
the result follows by choosing $c$ sufficiently large. [Note that first we choose $c$ sufficiently large, then we let $n \to \infty$.]

Update: This argument only holds for some cases. See italicized additions below.

Let $S_{n,k}$ denote the number of permutations in which the first $k$ elements are not fixed. I published an expository paper on these numbers earlier this year. See “Deranged Exams,” (College Mathematics Journal, 41 (3): 197-202, 2010). Aravind’s formula is in the paper, as are several others involving $S_{n,k}$ and related numbers.

Theorem 7 (which I also mention in this recent math.SE question) is relevant to this question. It’s
$$S_{n+k,k} = \sum_{j=0}^n \binom{n}{j} D_{k+j},$$
where $D_n$ is the number of derangements on $n$ elements. See the paper for a simple combinatorial proof of this.

Since $D_n$ grows as $n!$ via $D_n = \frac{n!}{e} + O(1)$ (see Wikipedia’s page on the derangement numbers), and if $k$ is much larger than $n$,
the dominant terms in the probability
$\frac{S_{n+k,k}}{(n+k)!}$ are the $j = n$ and $j = n-1$ terms from the Theorem 7 expression. Thus we have
$$\frac{S_{n+k,k}}{(n+k)!} \approx \frac{D_{n+k} + n D_{n+k-1}}{(n+k)!} \approx \frac{1}{e}\left(1 + \frac{n}{n+k}\right) \approx e^{-1} e^{\frac{n}{n+k}} = e^\frac{-k}{n+k},$$
where the second-to-last step uses the first two terms in the Maclaurin series expansion for $e^x$.

Again, this argument holds only for (in my notation) $k$ much larger than $n$.

This is not an answer but an observation that the quantity is equal to $\sum_{i=0}^{k} \frac{{(-1)}^i}{n!}{k \choose i}(n-i)!$.

Continuing Aravind’s observation, write it like
$$1 – \frac{k}{n} + \frac{1}{2!} \left( \frac{k}{n} \right)^2 \frac{1-1/k}{1-1/n} – \frac{1}{3!} \left( \frac{k}{n} \right)^3 \frac{(1-1/k)(1-2/k)}{(1-1/n)(1-2/n)}$$
$$+ \dots + (-1)^k \frac{1}{k!} \left( \frac{k}{n} \right)^k \frac{(1-1/k)(1-2/k)\dots (1-(k-1)/k)}{(1-1/n)(1-2/n) \dots (1-(k-1)/n)}.$$
If $k$ and $n$ are large, this seems to be close to the $k$th degree Maclaurin polynomial
for $e^z$ evaluated at $z=-k/n$ (although I don’t have an estimate of the error).