No.18

Why does 0! = 1 ?

Usually n factorial is defined in the following way:

n! = 1*2*3*...*n

But this definition does not give a value for 0 factorial, so a natural question is: what is the value here of 0! ?

A first way to see that 0! = 1 is by working backward. We know that:

                1! = 1
     2! = 1!*2 
                2! = 2
     3! = 2!*3
                3! = 6
     4! = 3!*4
                4! = 24
             

We can turn this around:

                4! = 24
     3! = 4!/4
                3! = 6
     2! = 3!/3
                2! = 2
     1! = 2!/2
                1! = 1
     0! = 1!/1
                0! = 1

In this way a reasonable value for 0! can be found.

How can we fit 0! = 1 into a definition for n! ? Let's rewrite the usual definition with recurrence:

      1! = 1
      n! = n*(n-1)! for n > 1
      

Now it is simple to change the definition to include 0! :

      0! = 1
      n! = n*(n-1)! for n > 0
      

Why is it important to compute 0! ?

An important application of factorials is the computation of number combinations:

               n!
   C(n,k) = --------
            k!(n-k)!
         

C(n,k) is the number of combinations you can make of k objects out of a given set of n objects. We see that C(n,0) and C(n,n) should be equal to 1, but they require that 0! be used.

                      n!
   C(n,0) = C(n,n) = ----
                     n!0!

So 0! = 1 neatly fits what we expect C(n,0) and C(n,n) to be.

Can factorials also be computed for non-integer numbers? Yes, there is a famous function, the gamma function G(z), which extends factorials to real and even complex numbers. The definition of this function, however, is not simple:

          inf.
   G(z) = INT x^(z-1) e^(-x) dx
           0

Note that the extension of n! by G(z) is not what you might think: when n is a natural number, then G(n) = (n-1)!

The gamma function is undefined for zero and negative integers, from which we can conclude that factorials of negative integers do not exist.

(from Math Forum)