How can the factorial of 0 be 1?

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Answer 1 · 2015-06-10T15:50:14

If you know the value of $n!$ then you can calculate $(n-1)!$ as
$(n!)/n$ ; since $1! =1$ then $0! = (1-1)! = 1/1 = 1$

Actually Nelson's answer if probably correct, but there is some justification for the definition.

Answer 2 · 2015-06-10T15:58:16

Because there is one permutation of zero objects.

Interpreting $n!$ to be the number of permutations of $n$ objects.

And agreeing that there is a permutation of $0$ object (namely the empty permutation).

Leads one to state that $0! = 1$

Answer 3 · 2015-06-11T21:16:50

We can do some MATH and STUFF!

At $n_0 = 1$ :
$(n!)/((n+1)!) = (1*cancel(2*3*4*...*n))/(cancel(2*3*4*5*...*n*)(n+1))$

$= 1/(n+1)$

If $(n!)/((n+1)!) = 1/(n+1) = (0!)/(1!)$ , then, with $n = 0$ :

$0! = 1!*(1/(0+1)) = (1!)/(1) = 1$