2007-02-27 23:20:54 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Suprisingly, 0! is more easier calculated than imagined. Another way of looking at it is "in how many ways do you arrange nothing" Only in ONE way "Dont arrange".3 things could be arranged among themselves in 3!=6 ways. Eg ABC,ACB,BCA,BAC,CBA and CAB. 2 things could be arranged in 2!=2 ways. Eg AB and BA. 1 could be arranged in 1!=1 way." A". That's it! And 0 thing(or nothing) could be arranged as 'dont arrange anything'. Hence 0!=1
2007-02-28 02:00:26 · answer #1 · answered by Mau 3 · 1⤊ 0⤋
A factorial of a number x is:
x!=1*2*...*(x-1)*x (1)
So we have (x-1)!=x!/(x-1) (2)
Of course if x is zero then the formula (1) cannot be applied because the formula (1) starts from 1, so in order to have the formula (2) to be consistent for all the numbers including 0 we define 0!=1. (This is a definition which means it has no proof, we just say when we write 0! we mean 1)
2007-02-28 00:59:23 · answer #2 · answered by kartik 2 · 1⤊ 1⤋
A factorial of a number x is:
x!=1*2*...*(x-1)*x (1)
So we have (x-1)!=x!/(x-1) (2)
Of course if x is zero then the formula (1) cannot be applied because the formula (1) starts from 1, so in order to have the formula (2) to be consistent for all the numbers including 0 we define 0!=1. (This is a definition which means it has no proof, we just say when we write 0! we mean 1)
2007-02-27 23:29:54 · answer #3 · answered by costasgr43 2 · 1⤊ 1⤋
It's a convenience, just the same as "1 is not a prime number".
If 0! was something else (or undefined), then lots of quite elegant formulae would have to have special cases for zero. Allowing 0! to be 1 means that we can use the same formulae for the "zero" case. That's all.
2007-02-27 23:29:32 · answer #4 · answered by Anonymous · 2⤊ 0⤋
Because factorial can be defined recursively.
For a natural n n!=n*(n-1)!.
1 = 1! = 1*0!
Thus, 0! = 1.
And it works well with combinatorics.
2007-02-27 23:28:02 · answer #5 · answered by Amit Y 5 · 1⤊ 1⤋
n! = (n-1)!n
Putting n = 1 gives 1! = 0!
Therefore 0! = 1
2007-02-27 23:35:56 · answer #6 · answered by Anonymous · 2⤊ 0⤋
The product of no numbers is 1 (the multiplication identity) - just like the sum of no numbers is zero (the additive identity).
Look up http://en.wikipedia.org/wiki/Empty_product for this and other similar "empty product" issues such as 1^0 = 1.
2007-02-27 23:55:31 · answer #7 · answered by Anonymous · 1⤊ 0⤋
suppose n is an integer
n!=(n-1)!
substitute n=1
n!=(n-1)!
1!=0!
1=0!
0!=1
I think the purpose of 0!=1 is for us to be easy to solve problems using counting techniques like permutation and combination...
2007-02-27 23:28:21 · answer #8 · answered by Bored 3 · 0⤊ 3⤋