Is 0! = 1? why? give full details.?

Question

let n be any positive integer.
n! = n(n-1)(n-2)(n-3)...3*2*1
then why not 0! = 0(0-1)(0-2)(0-3)...
in the series constants like 1,2,3,... will not come because these are > 0.
secondly 1! is also 1 will it not mean that 0! = 1! or 0 = 1

Answer 1

THE FORMULA
n! = n(n-1)(n-2)(n-3)...3*2*1
IS CORRECT BUT CONTAINS SOME LIMITATIONS ALSO.
IT IS TRUE FOR NATURAL NOS. ONLY.
0! = 0(0-1)(0-2)(0-3)...
THIS IS NOT TRUE
0! IS DEFINED AS 1
WHICH NEVER MEANS 0 = 1
AS CANCELLATION OPERATIONS INVOLVING 0 ARE NOT ALLOWED.
AS AN INSTANCE
4^0 = 1 AND 678^0 IS ALSO 1
WILL IT MEAN 4 = 678 OR 2 = 339
DO YOU AGREE? THINK.
WHY 0! = 1 IS A TRICKY QUESTION AND NOT EVEN A MATHEMATICIAN KNOW THE RIGHT ANSWER. HAVE YOU ASKED IT TO YOUR MATHEMATICS TEACHER? I AM SURE THAT HE / SHE ALSO CANNOT GIVE THE RIGHT ANSWER.

Answer 2

0! is defined to be 1. That's the reason.

0! = 1! doesn't mean that 0 = 1. You can't just divide out the factorial operator.

Answer 3

Actually, the proper definition of a factorial is:
0! = 1, and
n! = n(n - 1)(n - 2) .... (2)(1)

The reason why 0! is 1 is because it fits with the recursive definition. That is, we can express n! as n(n - 1)!

That means 1! can be expressed as 1(0)! = 1.

Answer 4

you can find the factorial for +ive integers only

now let us take

2! =2(2-1)
= 2*1
= 2

1! = (1-1)
= 0
is it true no this not

now consider this

3! = 1* 2*3
2! = 1*2
1! =1
0! =1

this is why 0! =1

Answer 5

one 0

Answer 6

Argue about it all you want to, and any way you want to. If you can do more consistent mathematics, using some different value for 0!, than professional mathematicians can do with 0! = 1, then you win and they lose. Otherwise, you lose.

Answer 7

its one... the same reason why 2^0 = 1

Answer 8

It is defined as such:

The special case 0! is defined to have value 1, consistent with the combinatorial interpretation of there being exactly one way to arrange zero objects (i.e., there is a single permutation of zero elements, namely the empty set 0).

Answer 9

3y + 2Z = 12 y - z = 9. from the 2d: y = z+9 substitute into the 1st 3(z+9) + 2z = 12 3z + 27 + 2z = 12 5z + 27 = 12 subtract 27 from the two sides 5z = -15 divide by potential of employing 5 z = -3 y = z+9 = -3+9 = 6 the respond is C.

Answer 10

3! = 3*2*1
2! = 2*1
1! = 1
0! = the empty product (the product of no numbers). This is equal to one. Read more about it here: http://en.wikipedia.org/wiki/Empty_product

If you find it hard to swallow, then ask yourself why you swallow this (same reasoning)...
x^3 = x*x*x
x^2 = x*x
x^1 = x
x^0 = the empty product = 1.