hey why is 0! = 1?

Question

I have always wanted to know why 0! (0 factorial) is equal to 1 anyone know?

I hope you know that factorial !
for example
3! = 3 x 2 x 1 = 6
and 5! = 5 x 4x 3 x 2 x 1 = 120
and Y! = Y x ( Y-1) x (Y-2) x........x 1
so why is 0! =1

Answer 1

There is a general principle involved here. n! is originally defined as the product of all intgers from 1 to n. Since 0! is not defined you can define it any way you want, although it is probably a good idea to define 0! = 1.

Being greedy I could define factorial on the fractions. The original definition is not much help with 2/3! Someone has done this and called it the gamma function.

We define a^n (n is a positive integer) as multiplying a times itself n times.

but what does a ^1/2 mean. I can't multiply 'a' times itself a half of a time; it's not defined. So I define it using roots and integral powers.

n x m is defined as adding m to itself n times.

But what about 3/2 x m? Can I add m to itself 1 and 1/2 times?
I fix this uncertainty by defining what it means to multiply a rational number times an integer.

What about a > 0 and a^pi? I defined a^(n/m) as (a^1/m)^n, that is in terms of roots and powers.

But pi isn't a fraction. Not to worry, it's eventually defined.

Some of these these things are discussed at
mathematicsteacher.org

There is also a link to "The Calculus: An Opinion" where these are mentioned also.

Answer 2

The formula for choosing k items from n things is:

n!/k!(n-k)!.

For example, the number of handshakes when everyone in a group of 5 people shakes hands is:

n = 5 (Five people)
k = 2 (2 People per handshake)

Hence,

5!/(2! 3!) = 10.

However, if there were only 2 people and everybody shakes hands with everybody else, there is only one handshake.

n = 2 (2 people)
k = 2 (2 people per handshake)

We get,

2! / (2! 0!) = 1

So, clearly 0! has to be 1.

Hence, we have to define 0! = 1.

Answer 3

0! = 1
as an instance of the convention that the product of no numbers at all is 1. This fact for factorials is useful, because the recursive relation (n + 1)! = n! × (n + 1) works for n = 0, and
this definition makes many identities in combinatorics valid for zero sizes.

Answer 4

You know n!(Factorial) = n+1^1/2.
So, 0!= 0+1^1/2=1

Answer 5

In permutation/combination problems 0! has to be =1.
If you have N items (such as 52 cards in a deck) and you remove x items(5 cards for a poker hand) and the order that they were removed does not matter(5 cards in a poker hand can be dealt in any order) you can calculate the number of possible outcomes(number of possible hands)

Possible outcomes = N!/( (x!)*(N-x)! )
so there are (52!)/ (5!*47!) = 2,598,960 possible poker hands.

But if you deal the entire deck, there is only 1 possible way to do this.

52!/( 52!*0!) becomes 1/(1*0!) = 1/0!
Number of possible ways to deal 52 cards = 1.
So 1 = 1/0!
solving for 0! gives 0! =1

Answer 6

Check now.

k!* (k+1) = (k+1)!
=>
0! 1 = 1! = 1
0! = 1

Answer 7

It is a definition. You should not think of it as a pattern. It was defined this way in order to make many mathematical formulas easier to create without having to consider the case of one of the variables being 0.

Answer 8

My math teacher told us to think about it w/ ice cream cones. He said, "How many different arrangements are there of n flavors in an ice cream cone?" So if n = 3, there would be 3! = 6 permutations. If n=0, there is only ONE possibility that would work: a cone with no scoops of ice cream in it.

Answer 9

In mathematics, an empty product, or nullary product, is the result of multiplying no numbers. Its numerical value is 1, the multiplicative identity, just as the empty sum — the sum of no numbers — is zero, or the additive identity

Answer 10

The ! next to the zero would have to be a 1