Why does $\frac{\displaystyle\sum_{x=3}^n{\frac{n!}{(n-x)!}}}{n!}\text{ as }n\to\infty$ give euler's number?

Question

TL; DR
While wondering how many permutations are there to fit n characters in 3+4+5+6+...+x character words (Like in the game Word Cookies), dividing by n! give e
This:

$$\frac{\displaystyle{\sum_{x=3}^n{\frac{n!}{(n-x)!}}}}{n!}\text{ as }n\to\infty$$

Gives e (2.718281828) somehow.


Here is the full story:

I was playing Word Cookies (I suggest looking at the image for better understanding of the game)
and while programming a bot that would play the game for me,
I was wondering how many possible permutations are there for n characters
that need to fit inside words with 3-x characters (depends on the word, as shown in the picture).
At first, my friend came up with this (wrong) formula (trust me its relevant):

$\displaystyle\sum_{x=3}^n{n!}$

I Later found out this formula (right):

$\displaystyle\sum_{x=3}^n{\frac{n!}{(n-x)!}}$

Explanation of the formula:

Let's say I have 5 characters say a,b,c,d,e , how many possible ways are there to fit them
in a 2 character word?
(or, how many 2 character words are possible using the characters a,b,c,d,e)
The answer is
$\frac{5!}{(5-2)!}$
Now, how many possible words are there using a,b,c,d,e that fit in a 2 character word + 3 character word? Answer:
$\frac{5!}{(5-2)!}$ + $\frac{5!}{(5-3)!}$
So the sigma just sums them up.

Then he said that it makes sense that the right formula is bigger than the wrong one (with factorial).

So I decided to check how much is it bigger (the ratio) so, I wrote in my calculator:

$\frac{\displaystyle\sum_{x=3}^8{\frac{8!}{(8-x)!}}}{8!}$ (n=8)

And got the result: 2.716666
Immediately I thought of the golden ratio. So I checked bigger numbers seeing if I can get closer to e.
When n=15, the result is 2.718281828

SO. my question: anyone has an explanation for that? It's really interesting as to why this happens.

Answer 1

More generally,

$\begin{array}\\ s(m, n) &=\dfrac{\sum_{x=m}^n{\dfrac{n!}{(n-x)!}}}{n!}\\ &=\dfrac{n!\sum_{x=m}^n{\dfrac{1}{(n-x)!}}}{n!}\\ &=\dfrac{\sum_{x=m}^n{\dfrac{1}{(n-x)!}}}{1}\\ &=\sum_{x=m}^n{\dfrac{1}{(n-x)!}}\\ &=\sum_{x=0}^{n-m}{\dfrac{1}{x!}}\\ &\to\sum_{x=0}^{\infty}{\dfrac{1}{x!}} \qquad\text{as } n \to \infty\\ &=e\\ \end{array} $