Factorial ($n!$)

For a nonnegative integer $n$, the factorial $n!$ is the product of the positive integers from $1$ to $n$, with the empty product equal to $1$:Thus $1!=1$, $2!=2$, $3!=6$, $4!=24$ and $5!=120$. The factorial counts the permutations of $n$ distinct objects: arranging $A,B,C$ gives exactly $ABC,ACB,BAC,BCA,CAB,CBA$. The choice tree in permutation shows the three choices for the first position, two for the second, and one for the last.

The value $0!=1$ has a counting explanation: the empty set has exactly one ordering, the empty list. Equivalently, it has exactly one bijection to itself, the empty function. There are no choices to make, and one completed outcome. The factorial therefore obeys the recurrence relationThe useful distinction is how many positions are filled. For integers $0\le k\le n$, choosing and ordering $k$ distinct objects by sampling without replacement gives the falling factorial, while choosing an unordered $k$-element subset gives the binomial coefficient:For five people, assigning captain and deputy gives $5!/3!=20$ outcomes; choosing a two-person team gives $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$. Every team has exactly $2!$ assignments of the two roles, explaining the division. When $k=0$, both counts are one. The factorial is useful here because one count of full orderings yields both formulas by cancellation and by forgetting order. ^[ref]

Distinctness matters. If the two copies of $A$ in $AAB$ are indistinguishable, the visible arrangements are just $AAB,ABA,BAA$, numbering $3!/2!=3$. Temporarily labelling them $A_1,A_2$ gives six permutations, but exchanging the two labels leaves each visible word unchanged, so every word was counted $2!$ times.

Proof outline. For $n\ge 1$, choose the first object in $n$ ways, then order the remaining $n-1$ objects in $(n-1)!$ ways. The rule of product gives the recurrence relation; repeated application gives the displayed product, starting from the one empty ordering. This explains why the factorial grows quickly: every additional distinct object introduces another multiplicative factor.

The double factorial multiplies the positive integers of the same parity as $n$ up to $n$: $(2m-1)!!=1\cdot 3\cdots (2m-1)$ and $(2m)!!=2\cdot 4\cdots 2m$. Set $(-1)!!=0!!=1$. In particular, $(2m-1)!!=(2m)!/(2^{m}m!)$, the number of pairings of $2m$ labelled objects.

The falling factorial is the product of $j$ successive descending factors, with $(t{)}_0=1$. Another notation is $t^{\underline{j}}$; this article uses $(t{)}_j$ for descending factors, whereas the rising factorial uses the same notation for ascending factors. For a nonnegative integer $t$, it counts ordered selections of $j$ distinct objects and vanishes when $j>t$. Powers expand as $t^j={\sum }_{\ell =0}^{j}S(j,\ell )(t{)}_{\ell }$, where $S(j,\ell )$ is a Stirling number of the second kind.

For a nonnegative integer $n$, the rising factorial is the product of $n$ consecutive increments starting at $a$, with $(a{)}_0=1$. Another notation is $a^{\overline{n}}$; the falling factorial uses descending factors. It compactly expresses the coefficient recurrence of a confluent hypergeometric function of the first kind.