# Factorial

Factorial is most simply defined as a function on positive integers. 5 factorial (written as $$5!$$) means $$1*2*3*4*5$$. In general then, for a positive integer $$n$$, $$n!=\prod_{i=1}^{n}i$$. For applications to combinatorics, it will also be useful to define $$0! = 1$$.

## Applications to Combinatorics

$$n!$$ is the number of possible orders for a set of $$n$$ objects. For example, if we arrange the letters $$A$$, $$B$$, and $$C$$, here are all the options:

$$ABC$$
$$ACB$$
$$BAC$$
$$BCA$$
$$CAB$$
$$CBA$$
You can see that there are $$6$$ possible orders for $$3$$ objects, and $$6 = 3*2*1 = 3!$$. Why does this work? We can prove this by induction. First, we’ll see pretty easily that it works for $$1$$ object, and then we can show that if it works for $$n$$ objects, it will work for $$n+1$$. Here’s the case for $$1$$ object.
$$A$$
$$1 = \prod_{i=1}^{1}i = 1!$$
Now we have the objects $$\{A_{1},A_{2},...,A_{n},A_{n+1}\}$$, and $$n+1$$ slots to put them in. If $$A_{n+1}$$ is in the first slot, now we’re ordering $$n$$ remaining objects in $$n$$ remaining slots, and by our induction hypothesis, there are $$n!$$ ways to do this. Now let’s suppose $$A_{n+1}$$ is in the second slot. Any orderings that result from this will be completely unique from the orderings where $$A_{n+1}$$ was in the first slot. Again, there are $$n$$ remaining slots, and $$n$$ remaining objects to put in them, in an arbitrary order. There are another $$n!$$ possible orderings. We can put $$A_{n+1}$$ in each slot, one by one, and generate another $$n!$$ orderings, all of which are unique, and by the end, we will have every possible ordering. We know we haven’t missed any because $$A_{n+1}$$ has to be somewhere. The total number of orderings we get is $$n!*(n+1)$$, which equals $$(n+1)!$$.

## Extrapolating to Real Numbers

The factorial function can be defined in a different way so that it is defined for all real numbers (and in fact for complex numbers too).

We define $$x!$$ as follows:
$$x! = \Gamma (x+1),$$
where $$\Gamma$$ is the gamma function:
$$\Gamma(x)=\int_{0}^{\infty}t^{x-1}e^{-t}\mathrm{d} t$$
Why does this correspond to the factorial function as defined previously? We can prove by induction that for all positive integers $$x$$:
$$\prod_{i=1}^{x}i = \int_{0}^{\infty}t^{x}e^{-t}\mathrm{d} t$$
First, we verify for the case where $$x=1$$. Indeed:
$$\prod_{i=1}^{1}i = \int_{0}^{\infty}t^{1}e^{-t}\mathrm{d} t$$
$$1=1$$
Now we suppose that the equality holds for a given $$x$$:
$$\prod_{i=1}^{x}i = \int_{0}^{\infty}t^{x}e^{-t}\mathrm{d} t$$
and try to prove that it holds for $$x + 1$$:
$$\prod_{i=1}^{x+1}i = \int_{0}^{\infty}t^{x+1}e^{-t}\mathrm{d} t$$
We’ll start with the induction hypothesis, and manipulate until we get the equality for $$x+1$$.
$$\prod_{i=1}^{x}i = \int_{0}^{\infty}t^{x}e^{-t}\mathrm{d} t$$
$$(x+1)\prod_{i=1}^{x}i = (x+1)\int_{0}^{\infty}t^{x}e^{-t}\mathrm{d} t$$
$$\prod_{i=1}^{x+1}i = (x+1)\int_{0}^{\infty}t^{x}e^{-t}\mathrm{d} t$$
$$= 0+\int_{0}^{\infty}(x+1)t^{x}e^{-t}\mathrm{d} t$$
$$= \left (-t^{x+1}e^{-t}) \right]_{0}^{\infty}+\int_{0}^{\infty}(x+1)t^{x}e^{-t}\mathrm{d} t$$
$$= \left (-t^{x+1}e^{-t}) \right]_{0}^{\infty}-\int_{0}^{\infty}(x+1)t^{x}(-e^{-t})\mathrm{d} t$$
By the product rule of integration:
$$=\int_{0}^{\infty}t^{x+1}e^{-t}\mathrm{d} t$$
This completes the proof by induction, and that’s why we can define factorials in terms of the gamma function.

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• Mathematics

Mathematics is the study of numbers and other ideal objects that can be described by axioms.

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