does anybody have the proof or know where the answer can be found for the following question
0!=1
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does anybody have the proof or know where the answer can be found for the following question
0!=1
I wasn't a math major but 0!=1 couldn't happen.
1!= 1 2!=2x1 3!=3x2x1 etc etc so since there is no subset of positive integers below zero, 0! like the square root of -1 would fall into the imaginary number category.
AFAIK, I don't think it has any proof. It's just an assumption to ensure that while calculating permutations and combinations (among other things), you don't get stuck with a 0 in the denominator.
Cheers,
cgkanchi
please see this site
www.newdream.net/~sage/old/numbers/fact.htm
it has the value for 0! but how it came i dont know
I'm sorry i'm no maths expert. 0=0 to me, to think otherwisemakes no sense.
By the way x=un-known quantity
Spurt=Drip under pressure
Of course 0!=1
(joke) :DCode:#include <stdio.h>
main()
{
if(0!=1)
printf("Hey! I'm right!");
else
printf("You Suck!");
return 0;
}