## anonymous 5 years ago why does is the integral form of n!(factorial)?

1. anonymous

wups, what is*

2. anonymous

The factorial function may be represented by a special function called, the gamma Function. You can't perform calculus on the factorial.

3. anonymous

Have you heard of that?

4. anonymous

Yes, but we are working with sequences/series, and are comparing the factorial with other integrals but he said that 0! = 1 and said to just believe him. Then he continued to write an integral form for n! which i believe is the gamma function.

5. anonymous

0! is 1 because we *define* it to be the case. It comes from permutations in combinatorics. If you have n objects and you're taking them r at a time, you have the total number of ways for doing this as,$^nP_r=n \times (n-1) \times (n-2) \times ...\times (n-r+1)$If we now multiply this expression by 1, but in the form,$1=\frac{(n-r)!}{(n-r)!}$we get$^nP_r=\frac{n \times (n-1) \times (n-2) \times ...\times (n-r+1)(n-r)!}{(n-r)!}$$=\frac{n!}{(n-r)!}$Now, if we have n objects and we want to arrange things in order n ways, we'd have$^nP_n=n!=\frac{n!}{(n-n)!}=\frac{n!}{0!}$For this to be true, we define$0!=1$

6. anonymous

Ok, so 1=0! rather than 0!=1. Think im gonna have to take some high level courses till i get to this but thanks, was interesting. You are very insightful.

7. anonymous

No probs.