does anybody have the proof or know where the answer can be found for the following question
0!=1

2. 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.

3. 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

www.newdream.net/~sage/old/numbers/fact.htm
it has the value for 0! but how it came i dont know

5. 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

6. Of course 0!=1

Code:
```#include &lt;stdio.h&gt;

main()
{
if(0!=1)
printf("Hey! I'm right!");
else
printf("You Suck!");

return 0;
}```
(joke)

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