0!=1!=!, and the question to you is...why?! (! is factorial, i.e. teh product of all the numbers up to that nubmer, like 5!=1*2*3*4*5 - but you all knew that, didn't you?)

Originally posted by Mouse2 n! is defined as follows:
0!=1
for n>0, n!=n*(n-1)!

which is basically the same thing as teh product of all the numbers up to that nubmer, but it explains the 0 case

don't you hate it when you post a puzzle that isn't so much a puzzle as a bit of knowledge that people'll get first time anyway? ah well-i've gotta find some more puzzle, or dig up some old uns...ah well-how about, for the mean time, integrate lnx?

Originally posted by genius 0!=1!=!, and the question to you is...why?! (! is factorial, i.e. teh product of all the numbers up to that nubmer, like 5!=1*2*3*4*5 - but you all knew that, didn't you?)

Why not just define k! as the number of permutations of k objects? If you have zero objects, then you have one possible permutation, so 0! = 1 with no special case definition.

I think those reasons are all made up after it was defined this way. That was probably because it fits best that 0!=1 and not that 0!=0, so people looked for 'logical' explanations. But that's the mathematicians way.

Originally posted by piderman I think those reasons are all made up after it was defined this way. That was probably because it fits best that 0!=1 and not that 0!=0, so people looked for 'logical' explanations. But that's the mathematicians way.

Originally posted by piderman I think those reasons are all made up after it was defined this way. That was probably because it fits best that 0!=1 and not that 0!=0, so people looked for 'logical' explanations. But that's the mathematicians way.

Even sillier is the following definition: n! := Gamma(n+1)

i think another reason 0!=1 is because if you end up with a 0! in the denominator of a fraction, then the whole sentence becomes undefined. and sometimes the are solutions. like n choose r -
5 choose 0-
5!/(5-0)!*0!
there is one way to choose 0 out of 5, so it is possible. if 0!=0, then it would be undefined, whereas now it is 1.

Originally posted by royalchicken Why not just define k! as the number of permutations of k objects? If you have zero objects, then you have one possible permutation, so 0! = 1 with no special case definition.