I have not found in any book by what procedure the value of zero factorial occurs. If there is any mathematical proof in any book please suggest it.
The value of 0! is 1, according to the convention for an empty product.
In mathematics, an empty product, or nullary product, is the result of multiplying no factors.
The proof that 0! = 1 depends on your definition of "!".
If your definition is n! is the product of the nonnegative integers less than or equal to n, while your definition of "product" incorporates the convention "an empty product equals 1" then the proof of 0!=1 is a two line proof applying the two lines of the definition, one by one.
https://en.wikipedia.org/wiki/Factorial
If you prefer a different definition then you will have to find a different proof, of course.
@Carlo: There are no permutations of a set containing no objects so the definition "number of permutations" is not a good definition. (A permutation is a function from the set to itself which is a bijection).
It follows from the definition of Gamma function and the fact that N! = Gamma(N+1).
See, for instance,p.255, Fig.6.1 in Handbook of Mathematical Functions, Abramovitz and Stegun, Dover, 9th printing,1972.
@Vasily: the "fact" that N! = Gamma(N+1) depends on what is the definition of the gamma function, and what is the definition of "factorial". So the relation you mention, though true, is not the proof.
@Bui: you are right too, but the fact you mention is also a consequence of the definition of n! (whatever that is), not a proof. Before we can give a proof of 0! = 1 we have to state the definition of "!". Depending on the definition, the proof might be different. There are several different conventional definitions. They are all equivalent, of course, so there is no conflict.
I am agree with Fedorov. You can easily prove it by using Gamma function.
Put n factorial then subtitute by 0 and 1 so you will find 0 factorial equal 1
No Proof can be given because it is defined like that. The special case is defined to have value, 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).
The value of 0! is 1, according to the convention for an empty product.
In mathematics, an empty product, or nullary product, is the result of multiplying no factors.
Define T(n)=(n-1)! as Gamma function
T(n+1)=nT(n) so n! = n.(n-1)!
Substituted n=1, we get 1! = 1.0! so 0!=1
By Definition of Combinations If you Try To find nCn using multiplication principle of counting, you will gat answer as n!.
Now apply formula of nCr the it comes out to be n!/0!
now by equality, you will get 0!=1
n!=n(n-1)!
why we are blind to see the other way
1=0!=(0)(-1)!=(0) )(-1) )(-2)… )(-n+1) )(-n)!=?
0!/0=(-1)n-1(n-1)!(-n)!
So what is zero?
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Thank you @Sher Baz. I think this is the only mathematical proof, if there is any other, I have not found yet.
Read the paper mentioned below for a different insight into this topic.
Article A simple demonstration of zero factorial equals one
n! = 1*2*3*..............(n-1)*n, therefore 1!= 1..................................................(1) also n! = n*(n-1)! and for n = 1 1! = 1*0! ..............................................(2) from (1) and (2) 0! = 1
Well, logically m!=n! implies m=n.
Consequently, 1!=0! should imply 1=0.
How could we resolve this illegal result by mathematical rationality?
Well, n!=г(n+1)=nг(n)
1=1!=г(1+1)=1г(1)=г(1)
1=0!=г(0+1)=0г(0)
To overcome such obstacle, one may said the expression
n!=г(n+1)
is not defined when n=0 : 0!=г(0+1)=0г(0)
Hence, so does its consequence 0!
0! is not defined
proof of 0! = 1
We know number of combinations of ‘r’ objects chosen from ‘n’ objects is nCr=n! /(r!(n-r)!) Also we know that number of combinations of ‘n’ objects chosen from ‘n’ objects is ‘1’ so
1=n! /(n!(n-n)!)Which implies
1=1/0 !
0!=1
Theres no proof, it's a contradiction and means zero by itself equals one, and math doesn't work.
In lay terms, there's just one (non-trivial) way to arrange an empty box of toys or a box containing one toy only.
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