The formula for “choosing ‘k’ permutations from a set of ‘n’ elements’ = P(n, k) = (n!)/( (n-k)! ). For set ‘C’, this was 3!/1! = 6/1 = 6.

Finally for this section, the number of “k-subsets” also is calculated with factorials. This is the problem for combining selected elements but without regard to permuting them into various sequences. From set ‘C’ above, the three 2-subsets are {a,b}, {a,c}, {b,c}.

The formula for combining or “choosing” k-subsets is usually shown in the above image, where C(n, k) = (n!)/( k! * (n-k)! ).

These concepts and formulas are useful in discrete probabilty theory, where one must determine the proportion of items that might be selected versus all the combinations that could be chosen.

## Extending the Factorial Beyond Positive Integers

The first and simplest extension is to define “zero factorial” = 0! = 1, for the reason that there is *exactly one way to arrange zero objects*. In other words, the empty set, {}, has exactly that one permutation.

The gamma function **for positive integers** is defined as Γ(n) = (n-1)!. This is *not* an extension of the factorial function, but simply defines part of the gamma function in terms of factorials.

The gamma function may be extended to all positive Real numbers by defining it as the smooth line that fits the points defined by the factorials of positive integers. This integration function plots a smooth curve that joins the (n, n!) points on the Cartesian plane for ( x > zero, Γ(x) ) in the Real numbers. Note the U-shaped curve in the upper right quadrant of the plot. In the integration version of the gamma function, ‘e’ is the base of natural logarithms, with a value of approximately 2.718.

Finally, the general gamma function extends to negative Real numbers and complex numbers. (Complex numbers have the form “z = x + iy”, where ‘z’ is Complex, ‘x’ and ‘y’ are Real, and ‘i’ is the square root of -1). The value of the integral function approaches infinity as the argument approaches each negative integer. Γ(x) never has the value zero.

Another equation defining the gamma function is Γ(x) = (x-1) * Γ(x-1), which is more reminiscent of the recursive definition of the factorial function on positive integers.

The gamma function has applications in probability and statistics, particularly when dealing with continous variables. However, any further discussion about the gamma function will be deferred for much later articles.

## A Future for Factorials in Decoded Science

We will also put an exclamation point on factorials through topics in permutations, combinations, and probability, as well as practical applications of each. To whet your appetite, one application of factorials is to calculate how many different bingo cards are required for a complete set.

**References**:

Weisstein, E. W. *Factorial; k-Subset; Permutation*. MathWorld–A Wolfram Web Resource. (1999-2012). Accessed August 8, 2012.

stanley ness says

You elitests talk past the layman who wants a simple explanation when used in a formula.

Ex.cubic inches of a pipe to calculate volume of water.

e says

I wrote javascript in the console to play with the concept.

to be honest, I think the reason math is so impenetrable for people is that if you took the same way people write math problems out to code, people would hate your guts. Everything is horrible shorthand. I came to this page because I wanted to know how the formula on this page works (http://www.homeschoolmath.net/teaching/sine_calculator.php).

There’s no explanation of the logic behind these concepts, or why they get you to the right answer. I can appreciate the quick and dirty need to just utilize functions and things people have already figured out, but for the curious mind there is so little nourishment. Why does it work, what do these sub calculations represent, if I wanted to recombined them in different ways to solve new problems, I can’t because I don’t know what they mean.

I can really struggle thinking just factorial out for an hour, to say, well the reason 3*2 works is that the 2 represents the number of times each element can appear at the beginning of the set, and 4*3*2 works because the each element can appear at the beginning of the set 3 times, and when you’ve accounted for those possibilities, the number of remaining variations for those sets are the number of times each element of the remaining elements are going to appear first, but be damned if anyone is going to try to put that in English, or spend more than the 5 minutes I spent hear trying to make it actually coherent to someone just encountering this concept. And the only reason I can sort of make sense of it is because I spent a lot of time thinking about it one time as a kid, not because any teacher tried to explain or even knew themselves why it worked.

ok that’s my rant ! At least the page I found above was better than the broken responses you normally find when asking ‘what actually is sin’ and people just tell you what sin is useful for.

var factorial = function(n) {

if (n > 2) { return n * factorial(n – 1) } else { return n } }

undefined

factorial(2)

2

factorial(3)

6

factorial(4)

24

factorial(5)

120

factorial(56)

7.109985878048635e+74

Seth says

Simplify: 5! 8!2!

K.S.Patel says

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

anna says

Woooooooow understood none of that ^.^

Jayden says

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

Randoman Lolmao says

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

Lee Shafer says

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

Poorva says

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

Bill says

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

Mike DeHaan says

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!)

joe franco says

why is 0! = 1

meshfields_blog says

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!. ■

Vincent Summers says

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