Interpret This Function First!

If we define $f(0) = 1$ and $f(1) = 0$ , then the recurrence relation for $f(n)$ can be rewritten as $n! \; = \; \sum_{r=0}^n \binom{n}{r}f(n-r) \qquad \qquad n \ge 0$ and hence $1 \; =\; \sum_{r=0}^n \frac{1}{r!} \times \frac{f(n-r)}{(n-r)!} \qquad \qquad n \ge 0 \;.$ If we consider the generating function of the coefficients $f(n)$ defined by $F(x) \; = \; \sum_{n=0}^\infty \frac{f(n)}{n!} x^n$ then the above recurrence relation can be read (compare the coefficients of $x^n$ ) as $\frac{1}{1-x} \; = \; e^x F(x)$ and hence $F(x) \; = \; \frac{1}{1-x}e^{-x} \qquad \qquad |x| < 1$ so that $\frac{f(n)}{n!} \; = \; \sum_{r=0}^n \frac{(-1)^r}{r!}$ which makes $m = e^{-1}$ and the answer $\boxed{367}$ .

The following solution is intuitive and not very mathematically rigorous but i believe it is correct.

If there are $n$ distinct letters and $n$ corresponding envelopes, the given function can be thought of to represent the number of different ways the letters can be placed in the envelopes (one letter in each envelope) such that no letter is placed in the correct envelope. If $\left( \begin{matrix} n \\ r \end{matrix} \right) f(n-r)$ is assumed to represent the number of ways $r$ letters can be selected and placed in their correct envelopes and every other letter in their wrong envelope, then $\left( 1+\sum _{ r=1 }^{ n-2 }{ \left( \begin{matrix} n \\ r \end{matrix} \right) f(n-r) } \right)$ will be the number of ways one or more letters are placed in their correct envelopes, and the function will give the number of ways no letter is placed in the correct envelope, which matches with our initial assumption.

The function can therefore be alternatively written as : $f(n) =n!\left[ 1-\frac { 1 }{ 1! } +\frac { 1 }{ 2! } -\frac { 1 }{ 3! } +\cdots +\frac { { (-1) }^{ n } }{ n! } \right]$

$\therefore \frac { f(n) }{ n! } =\left[ 1-\frac { 1 }{ 1! } +\frac { 1 }{ 2! } -\frac { 1 }{ 3! } +\cdots +\frac { { (-1) }^{ n } }{ n! } \right]$

$\boxed{\underset { n\rightarrow \infty }{ lim } \frac { f(n) }{ n! } =\frac{1}{e}}$

We can define $f(0) = 1$ and $f(1) = 0$ , so that,

$\sum_{r=0}^{n} \dbinom{n}{r} f(n-r) = n!$

Substitute $r \mapsto n-r$ to get,

$\sum_{r=0}^{n} \dbinom{n}{r} f(r) = n!$

Now, consider $\text{S} = \sum_{k=0}^{n} (-1)^k \dbinom{n}{k} k!$

$\implies \text{S} = \sum_{k=0}^{n} \sum_{r=0}^{k} (-1)^k \dbinom{n}{k} \dbinom{k}{r} f(r)$

$= \sum_{k=0}^{n} \sum_{r=0}^{k} (-1)^k \dbinom{n}{r} \dbinom{n-r}{k-r} f(r)$

$= \sum_{r=0}^{n} (-1)^r \sum_{k=r}^{n} (-1)^{k-r} \dbinom{n}{r} \dbinom{n-r}{k-r} f(r)$

$= \sum_{r=0}^{n} (-1)^r \dbinom{n}{r} f(r) \sum_{k=r}^{n} (-1)^{k-r} \dbinom{n-r}{k-r}$

Note that the inner sum vanishes unless $n=r$ , thus,

$\text{S} = (-1)^r \dbinom{n}{r} \dbinom{n-r}{n-r} f(r) \quad ; \quad r=n$

$= (-1)^n f(n)$

$\implies f(n) = \sum_{k=0}^{n} (-1)^{n-k} \dbinom{n}{k} k!$

Substitute $k \mapsto n-k$ ,

$\implies f(n) = \sum_{k=0}^{n} (-1)^{k} \dbinom{n}{k} (n-k)! = n! \sum_{k=0}^{n} \dfrac{(-1)^{k}}{k!}$

So that $\displaystyle \lim_{n \to \infty} \dfrac{f(n)}{n!} = \dfrac{1}{e}$ making the answer $\boxed{367}$

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In general,

$a_{n} = \sum_{r=0}^{n} \dbinom{n}{r} b_{r} \iff b_{n} = \sum_{r=0}^{n} (-1)^{n-r} \dbinom{n}{r} a_{r}$