What is a “factorial,” and how is it used in mathematics?

## The Mathematical Definition of Factorial for Positive Integers

In mathematics, a factorial is a function applied to natural numbers greater than zero. The symbol for the factorial function is an exclamation mark after a number, like this: 2!

The usual definition of “n factorial” is “n! = n * (n – 1) * (n – 2) *…* 2 * 1,” where ‘n’ is a positive integer.

The first handful of factorial values from positive integers are: 1! = 1; 2! = 2; 3! = 3*2 = 6; 4! = 4*3*2 = 24; and 5! = 5*4*3*2 = 120. In the image, only the first five factorial results are plotted although the first ten have been calculated.

## Recursive Definition for the Factorial

A recursive definition for the factorial function is “n! = n * (n-1)!”, placing the lower limit for the recursion at n=2.

## Factorial Function: Practical Uses

Although the factorial function deals with repeated multiplication, its most obvious use in math is to compute the *number of ways in which ‘n’ objects can be permuted*.

A permutation is a re-arrangement of a set. For example, set A={a} has exactly one arrangement. However, set B={a,b} could be re-arranged as {b,a}; and set C={a,b,c} has the six permutations {(a,b,c), (a,c,b), (b,a,c), (b,c,a), (c,a,b), (c,b,a)}.

Note that |A| = 1 (set A has one element), |B| = 2 and |C| has three; but they have 1, 2 and 6 permutations respectively. Our readers are invited to list the permutations for set D={a,b,c,d}, but there should be two dozen in that permutation set.

This extends to selecting permutations of *an ordered subset*. From set C={a,b,c}, how many ways can two elements be selected and permuted? The solution set is {(a,b), (b,a), (a,c), (c,a), (b,c), (c,b)}, and has six elements.

**Click to Read Page Two: The Formula for Choosing K**

Simplify: 5! 8!2!

5!8!2! = 120*40320*2 = 9676800

5! = 5*4*3*2*1 = 120

8! = 8*7*6*5*4*3*2*1 = 40320

2! = 2*1 = 2

Woooooooow understood none of that ^.^

Can you do !!? For example, 6! Would be 6x4x2

No, I’m afraid that’s wrong. The correct term is 6•5•4•3•2.

I just saw the clock where all the hours are expressed as some formula involving only 9. I don’t get 7 o’clock. To make 7 you need that last part to =1. There is a line with .9 under it. I never had that in school.what is that?it obviously equals one. Lee

9 – Root 9 + .9 with bar

= 9 – 3 + .9

= 6 + .9

= 6.9

= 7

as per the maths rule, we sum up the number when it exceeds 0.5 .. Here the .9 has a bar .i.e. 6.999999999999 (Infinite) hence we summed it up to 7

6 point (infinite number of 9s) is the same as 7. No need to round up.

5 is wrong as it uses 9! (factorial) to mean 9+8+…+1 rather than multiplied by

Excellent question. By appeal to authority in Wolfram, “The special case 0! is defined to have value 0!=1, consistent with the combinatorial interpretation of there being exactly one way to arrange zero objects (i.e., there is a single permutation of zero elements, namely the empty set)”.

Wikipedia’s “Gamma function” has the line

“Combining this with Γ(1) = 1, we get Γ(n) = (n-1)!”

when defining the gamma function.

Plus…

SosMath discusses the gamma function as an integral, noting that “e^(-1) = 1″ (and should have noted “x^0 = 1″). Further down the page, they derive Γ(1) = 1 from the integral.

(See http://sosmath.com/calculus/improper/gamma/gamma.html ).

My basic answer is: it works out nicely when relating factorials to the gamma function.

(Sorry I hadn’t checked my mail over the weekend!)

why is 0! = 1

Because the number of possible permutations of 0! is 1. The empty set { } is still one empty set with one permutation inside.

Proof: Let n=1, using the definition n! = n*(n-1)! then is

1! = 1*0! which can be simplified to 1 = 0!. ■

As I get older, I begin to accept what some might call the fudge factor. If the fudge factor works, run with it.