Factorials

[carebear]

I don't understand why 0! and 1! are both equal to 1. Please help?

I understand that 1! is 1....but in my mind 0! should be 0 x 1 which would equal 0 not 1?!?!

 

[lookagain]

I see that for n!, at least for integers n >= 1, it is the number of ways that n items can be arranged
with regard to order. Then I see 0! as exactly equal to one way to order no items.

 

[Denis]

http://www.google.ca/#hl=en&source=hp&b ... e234543ea9

 

[galactus]

Take \(\displaystyle n!=n(n-1)!\)

Let n=1

We get

\(\displaystyle 1!=1(1-1)!=1\cdot 0!=1\)

Because we know there is 1 way to arrange 1 item.

It sure would make the counting and probability world more complicated if it were not equal to 1.

Now, if only we can change the laws to make \(\displaystyle (x+y)^{2}=x^{2}+y^{2}\)
:wink:

 

[Deleted member 4993]

Take \(\displaystyle n!=n(n-1)!\)

Let n=1

We get

\(\displaystyle 1!=1(1-1)!=1\cdot 0!=1\)

Because we know there is 1 way to arrange 1 item.

It sure would make the counting and probability world more complicated if it were not equal to 1.

Now, if only we can change the laws to make \(\displaystyle (x+y)^{2}=x^{2}+y^{2}\) ? x and/or y = 0
:wink: