The answer to 4!=24, Why is 0! = 1.?

Answer 1

The fact that 0! = 1 is true simply by definition. It's defined to be so.

The reason for this definition is based partly on the properties of the gamma function, which is the continuous analog of the factorial function, such that Γ(x) = (x-1)!
The value of the Gamma function at 1 (which would be 0!) is, in fact, 1.

Another reason for this definition is that it's very convenient for infinite sums involving factorials. For example, the sum

e^x = Σ(from 0 to inf) x^n / n!

would be much more awkward if 0! did not equal 1.

Answer 2

Sometimes in school they teach it as a "law" or say it's just "defined that way", but it's actually quite provable.

The number 1 is called the empty product, the same way that 0 is the empty sum. When you add 2+3, you begin with the empty sum, so you're actually adding 0+2+3. But nobody thinks about it that way because you get the same answer. When you compute 4!, you begin with the empty product and multiply by 4!. You're actually doing 1*1*2*3*4. Again, nobody thinks about it that way because you get the same answer.

Now let's do this for 0!. The factorial operation n! is defined as the product of all integers less than or equal to n and greater than 0. There are no such integers for 0, so 0! contains no factors at all. 0! is completely empty. Thus we compute 0! by starting with the empty product 1 and multiplying it by 0!, which means we are multiplying 1 by nothing at all. Hence the result is 1.

Here's the short version.
0! = 1*0! = 1*{there's nothing to multiply by} = 1.

Answer 3

The definition of n! (n-factorial) for positive integers and zero is:

n! = n x (n - 1) x (n - 2) x .... x (n - (n - 2)) x (1)

n-factorial does not include (n - n) as a factor because then all answers would always = 0.

One can quickly discover that this definition becomes confusing for any values lower than 2, especially 1! and 0!.

1! = n x (n - 1) x ... where n = 1, in following the rule that we never multiply by (n - n), we quickly check the next term in the expansion, this is (n - 1), but since n =1, then we simply stop at n! = n, or 1! = 1.

So, 1! is sort of a trivial waste of time, unless one needs to figure out 0!, then it becomes important!

If we look at n! we notice a relationship between n! and (n - 1)!

Mainly that { [ n! ] / [ (n - 1)! ] } always = n for all positive integers n.

Thus, by this rule we see that if n = 1,
1! / 0! = 1,
Which according to algebra means that,
1! / 1 = 0!,
Since I just showed that 1! must equal 1 if we follow logical rules for n-factorial,
then 1! / 1 = 1 / 1 = 0!,

For brevity's sake, I will not prove that for all positive integers n above zero, n / n = 1, just trust me that 1 / 1 = 1,

0! must equal 1, otherwise, the whole rule falls apart and factorials all become undefined. But we know that all factorials n! above 1 work (this can be proven by someone else past or present).

I hope this helps explain factorial convention and answers your question well.

Answer 4

In general, we define things so that they are useful.

We define 0! = 1 because it is useful. We could leave it undefined, but there are reasons to define it as 1, as follows:

(1) It allows the equation n! = n * (n-1)! to be true for n=1.
(2) There is a nice function gamma(x) which is defined on most of the complex numbers, such that gamma(n)=(n-1)!. It turns out that gamma(1)=1.
(3) In general, an 'empty' product is often defined as 1. For example, x^0 = 1 for all x (and we even define 0^0=1.) 0! can be thought of as an empty product. If A and B are two sequences, the the product(A) * product(B) = product(A,B), where (A,B) is the result of taking the two finite sequences in order. To make this rule continue to work if A or B is an empty sequence, we'd want product(empty seq)=1.

Answer 5

There are two answers.

The first is by definition which is a fancy way of saying because.

The second is the real answer and will only make sense if you know what the gamma function is. The gamma function is an integral that is parameterized by a parameter z. since I cant type greek symbols
gamma(z)=integral from 0 to infinity of dt * t^(z-1)*e(-t) and has the remarkable property for integers greater than or equal to 1
gamma(z)=(z-1)!
Now if you think about extending the definition of factorials, put z=1 as a parameter for the integral, do the definite integral you end up with gamma(1)=0!=1
you should also know that when z=negative integer, gamma(negative integer)=infinity

Answer 6

The quantity n! (pronounced "n factorial" or "factorial n") is defined as the product of the n integers from 1 to n. The basic reason why 0! equals 1 is that it's merely a product of 0 factors; Such an empty product must be equal to 1, just like a sum of zero terms (an empty sum) must be equal to 0. Let me explain:

The product of (n+1) factors is clearly equal to the product of the first n factors multiplied by the last one. This is "clear" to everybody when n is 2 or more. To make this work for n=1 we have to state that a "product" consisting of a single factor is equal to that factor. It follows (for n=0) that a product of zero factors multiplied by any number x must be equal to x. Therefore, the product of zero factors must be equal to 1. (The same reasoning for sums leads to the conclusion that a sum of zero terms is equal to 0, which is less shocking to most people than the corresponding result for empty products.)

Defining n! as a product of n factors (1,2, ... n) when n is nonzero thus implies that the only consistent definition of 0! is 0!=1.

Another "advanced" argument is to define factorials in term of the analytic Gamma function ( G ), whose properties also imply a value 0!=1.

Answer 7

The simple answer is because it's a very convenient definition that allows a lot of equations to be written whithout having to have 0 be treated as a 'special case'.

The real answer is because the Gamma Function is a generalization of the factorial function over the reals (since Γ(n) = nΓ(n-1)), and Γ(0) = 1. You'll run into the Gamma function somewhere around the end of your 2'nd year of Calculost ☺

Doug

Answer 8

The factorial function (!) is a way of working out how many possible permutations there are of different sized groups (ie. how many ways can you arrange different objects).

For example, if you have one red, an one blue marble, there are 2 ways you can arrange them (2! = 2x1 = 2). These ways are:
R B
B R

For three things, there are 6 ways (3! = 3x2x1 = 6), so if a green marble was added, you would have:
R B G
R G B
G R B
G B R
B R G
B G R

Now suppose you had no marbles, there is only one way you can arrange that, and that is to have ... nothing. There is no other way to order it, so there is only one. Therefore, 0!=1 as there is only one way you can arrange 0 objects.

Answer 9

Think of permutation. What does n! mean?
n! is actually the number of permutations (arrangements) of n distinct objects all at a time without repetition, i.e. nPn.

So 0! will mean that number of arrangements of 0 objects which can be done in only 1 way that is by not arranged.

Hence from definition
0! = 1

Answer 10

easy!
4! = 4*3!, right?
divide both sides by 4 which still works
4! / 4 = 3!, right?

so technically.....
n! = n*(n-1) !

so when n = 1.....

1! = 1*(1-1)!
1! = (1-1)!
1! = 0!

ultimately 0! = 1, works every time!! that's why its true..=)