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Why 0! is always equal to 1? Why 0! is always equal to 1?

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Old Wednesday, August 04, 2010
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Default Why 0! is always equal to 1?
why 0! is always equal to 1 ??
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Old Wednesday, August 04, 2010
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If n! is defined as the product of all positive integers from 1 to n, then:

1! = 1*1 = 1
2! = 1*2 = 2
3! = 1*2*3 = 6
4! = 1*2*3*4 = 24
5! = 1*2*3*4*5=120

Hence:

n! = 1*2*3*...*(n-2)*(n-1)*n
and so on.

In this way n! can also be expressed as n*(n-1)! .


Therefore, at n=1, using n! = n*(n-1)!

1! = 1*0!

Resultantly, 1 = 0!

Or:

n! = n*(n-1)!
n!/n = (n-1)!
If n=1 Then:

1!/1 = (1-1)!
1*1/1 = (0)!
1/1 = 0!
1=0!
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Last edited by Raz; Wednesday, August 04, 2010 at 10:52 PM.
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Factorial of a number is the product of all the positive integers below that integer to be asked...
that n ! = n(n-1)(n-2).........

we have 0! = 1

since zero is neither positive nor negative ... hence we always take
0!=1 ...

there are other proves in books but in those they put 1!= 1 .though 1! itself depends on 0! .. that is 1!= 1(1-1)! = 0! = 1
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