#2**+13 **

0!=1

I could just say that it is define that way which is certainly true.

or I could say that 0! is the number of ways that 0 items can be odered - that would be just one way.

Now

5C1 is the number of ways that 1 item can be chosen from 5 it equals $$^5C_1=\frac{5!}{1!(5-1)!}=\frac{5!}{1*4!}=5\\\\$$

5C0 is the number of ways that 0 item can be chosen from 5 it equals [the answer has to be 1 there is only one way of chosing nothing]

$$\\^5C_0=\frac{5!}{0!(5-0)!}=\frac{5!}{0!*5!}=\frac{1}{0!}\\\\

$ the only way this can work is if 0!=1$\\\\

so\\\\

0!=1$$

Melody
Apr 10, 2015

#1**+5 **

Here's an informal "proof"....

3! = 4! / 4

2! = 3! / 3

1! = 2!/ 2

So, by extension

0! = 1! / 1 = 1

CPhill
Apr 10, 2015

#2**+13 **

Best Answer

0!=1

I could just say that it is define that way which is certainly true.

or I could say that 0! is the number of ways that 0 items can be odered - that would be just one way.

Now

5C1 is the number of ways that 1 item can be chosen from 5 it equals $$^5C_1=\frac{5!}{1!(5-1)!}=\frac{5!}{1*4!}=5\\\\$$

5C0 is the number of ways that 0 item can be chosen from 5 it equals [the answer has to be 1 there is only one way of chosing nothing]

$$\\^5C_0=\frac{5!}{0!(5-0)!}=\frac{5!}{0!*5!}=\frac{1}{0!}\\\\

$ the only way this can work is if 0!=1$\\\\

so\\\\

0!=1$$

Melody
Apr 10, 2015