Why 1!(1 factorial) is 1 and 0!(0 factorial) is 1? Explain it.
Answers
Before we proceed I think its good idea to define Factorial first, here is what Wikipedia says:
It is a non-negative integer n, denoted by n!, is the product of all positive integer less than or equal to n.
n!=n∗(n−1)∗(n−2)∗(n−3)∗...3∗2∗1
However the recursive definition of factorial is of more use in this proof.
n!={1n∗(n−1)!n=1n≥0
Recursive definition of Factorial leads to one interesting way of expressing factorial numbers.
n!=(n+1)!(n+1)
This is valid since, as we expand (n+1)! from recursive definition, we can cancel (n+1) term from both numerator and denominator to get n! . Or we can even calculate factorial in numerator and then evaluate the division
For example,
5!=6!6=7206
4!=5!5=1205
3!=4!4=244
2!=3!3=63
1!=2!2=22
In a similar way, if we try to express 0! we get
0!=1!/1=1
And this ends our proof that 0!=1 .
This proof is one of many ways, where 0! leads to 1 . But this one is quite explanatory in itself.
There is one more interesting question I would like to share which was in my mind back in those days.
WHY DO WE NEED 0! ?
And the answer is computation of number of Combination.
(n/k)=n!/k!(n−k)!
when k=n
(n/n)=n!/n!∗0!
This is one of many application of 0! . But this example give a really good idea where we can make use of 0!=1 .