3 factorial (written as 3!) is the product of all positive integers from 1 to 3.
This means 3 factorial equals 6.
are commonly used in permutations, combinations, and probability calculations. For example, 3! represents the number of ways to arrange 3 distinct items, which is 6 possible arrangements.
ELI5: What is a factorial and how does it work
What is this kind of factorial called?
What does 3 factorial mean?
sequences and series - $\sum k! = 1! +2! +3! + \cdots + n!$ ,is there a generic formula for this? - Mathematics Stack Exchange
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Is there another name for 52 factorial factorial factorial
So I know that 3 factorial is 123 132 213 231 312 321 But what about if you can use the same number twice? So like this: 111 112 113 121 122 123 131 132 etc.
In addition to the special functions given by J.M., an asymptotic expansion can be computed $$ \begin{align} \sum_{k=0}^n k! &=n!\left(\frac11+\frac1n+\frac1{n(n-1)}+\frac1{n(n-1)(n-2)}+\dots\right)\\ &=n!\left(1+\frac1n+\frac1{n^2}+\frac2{n^3}+\frac5{n^4}+\frac{15}{n^5}+O\left(\frac1{n^6}\right)\right)\\ &=\sqrt{2\pi n}\frac{n^n}{e^n}\left(1+\frac{13}{12n}+\frac{313}{288n^2}+\frac{108041}{51840n^3}+\frac{12857717}{2488320n^4}+O\left(\frac1{n^5}\right)\right) \end{align} $$ As with most asymptotic expansions, the series does not converge, and cannot be used to get an exact answer, but it gives a good approximation.
Edit: I forgot to give $$ \begin{align} \sum_{k=0}^nk!^2 &=n!^2\left(\frac11+\frac1{n^2}+\frac1{n^2(n-1)^2}+\frac1{n^2(n-1)^2(n-2)^2}+\dots\right)\\ &=n!^2\left(1+\frac1{n^2}+\frac1{n^4}+\frac2{n^5}+\frac4{n^6}+\frac{10}{n^7}+O\left(\frac1{n^8}\right)\right)\\ &=2\pi\frac{n^{2n+1}}{e^{2n}}\left(1+\frac1{6n}+\frac{73}{72n^2}+\frac{1049}{6480n^3}+\frac{157541}{155520n^4}+O\left(\frac1{n^5}\right)\right) \end{align} $$
(Too long for a comment)
I don't know if there's a simpler form, but the sum of factorials has certainly been well-studied. In the literature, it is referred to as either the left factorial (though this term is also used for the more common subfactorial) or the Kurepa function (after the Balkan mathematician Đuro Kurepa).
In particular, for $K(n)=\sum\limits_{j=0}^{n-1}j!$ (using the notation $K(n)$ after Kurepa), we have as an analytic continuation the integral representation
$$K(z)=\int_0^\infty \exp(-t)\frac{t^z-1}{t-1}\mathrm dt,\quad \Re z>0$$
and a further continuation to the left half-plane is possible from the functional equation $K(z)-K(z-1)=\Gamma(z)$
An expression in terms of "more usual" special functions, equivalent to the one in Shaktal's comment, is
$$K(z)=\frac1{e}\left(\Gamma(z+1) E_{z+1}(-1)+\mathrm{Ei}(1)+\pi i\right)$$
where $E_p(z)$ and $\mathrm{Ei}(z)$ are the exponential integrals.
The sum of squares of factorials does not seem to have a simple closed form, but the sequence is listed in the OEIS. One can, however, derive an integral representation that could probably be used as a starting point for analytic continuation. In particular, we have
$$\sum_{j=0}^{n-1}(j!)^2=2\int_0^\infty \frac{t^n-1}{t-1} K_0(2\sqrt t)\mathrm dt$$
where $K_0(z)$ is the modified Bessel function of the second kind.