why is the factorial of 0, 1?? 0!=1??

Answer 1

n! = n(n-1)(n-2)(n-3)...3x2x1
= n[ (n-1)(n-2)(n-3)...3x2x1]
= n (n-1)!

n!= n(n-1)!

(n-1)! = n! / n

substituting n = 1 gives

(1-1)! = 1/1

which means 0! = 1

Answer 2

It's defined as 1. Some people long long ago decided that's what it should be.

Also, factorial represents number of sets.

1! = 1 = { 1 }
2! = 2 = { 1, 2 } { 2, 1 }
3! = 6 = { 1, 2, 3 } { 1, 3, 2 } { 2, 1, 3 } { 2, 3, 1} { 3, 1, 2 } { 3, 2, 1}

0! has one set. What set you may ask? The empty (or 'null') set.
There's no ASCII character for it, but it looks like a big O with a slash going through it.

Answer 3

though there is no solid proof of 0!=1
there is a 'kaam-chalaau' proof :-

That equation (n!=n(n-1)!) just dictated to us where to put the parentheses. By making n=1, we can find 0!:






1!=1(0!)

hence
0! = 1! / 1

hence
0! = 1 / 1

hence
0!=1






And, it turns out that 0!=1 works very well in many situations (in probability, for example).
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The above proof that 0!=1 is based upon n!=n(n-1)!, which is in turn based upon the definition of factorial. So, it would seem to be a valid proof. But, 0! cannot be defined directly from the definition of factorial. So, mathematicians like to define 0! as 1, without proving it. So, the proof just amounts to a demonstration that defining 0! as 1 is consistent with the definition of factorial.

Answer 4

The factorial function is formally defined by
n! = n(n – 1)(n – 2) … (2)(1) for any positive integer n.
Recursively, it is defined as n! = n(n-1)!. Morover, 0! = 1.

The function that "fills in" the values of the factorial between the integers is called the Gamma function, denoted Γ(z) and for z! = 0, − 1, − 2,... defined by

Γ(z) = ∫ t^(z-1) exp(-t) dt from 0 to ∞.

Note that the Gamma function generalizes the definition of factorial. Its properties can be viewed on the site http://en.wikipedia.org/wiki/Factorial. Some of its properties are as follows:

(a)Γ(z + 1) = z Γ(z)
(b)Γ(z + 1) = z!

Trivially,

Γ(1) = 0! = ∫ t^(1-1) exp(-t) dt from 0 to ∞
= ∫ exp(-t) dt from 0 to ∞
= –exp(-t) evaluated from 0 to ∞
= [lim –exp(-t), t → ∞ ] – [–exp(-0)]
= 0 – [–1]
= 1.

This explains that Γ(1) = 0! = 1.

Answer 5

we can see the sequence here...
4! = 5!/5 = 120/5 = 24.
3! = 4!/4 = 24/4 = 6
2! = 3!/3 = 6/3 = 2
1! = 2!/2 = 2/2 = 1
0! = 1!/1 = 1/1 = 1
The sequence gives 0!=1

Answer 6

We sometimes see that 0!=1.why is that?It would seem that we cannot do the multiplications from one "up to" zero. And, if we go backwards (1x0), we get zero.

Experimenting with factorials, we come up with n!=n(n-1)!. For example 17!=17x(16!):

16!=1x2x...x16
17!=(1x2x...x16)x17

That equation (n!=n(n-1)!) just dictated to us where to put the parentheses. By making n=1, we can find 0!:

1!=1(0!)
0!=1

And, it turns out that 0!=1 works very well in many situations (in probability, for example).

Answer 7

I think it's just defined that way. 0!=1

Answer 8

we know that n! = n.(n-1)!
now 1! = 1 ............(a)
also 1! = 1.(1-1)!
since, 1-1 = 0
so, 1! = 1.0!
but 1! = 1 from (a)
so, 1 = 1.0!
and hence
0! = 1

Answer 9

Abhishek has given a simple and easy to follow proof. Other after him are saying the similar thing.

Answer 10

cant say
but it a constant kinda thing that u need to assume before applying it