2006-12-08 03:16:33 · 9 answers · asked by JAMES 4 in Science & Mathematics ➔ Mathematics
There's no "formula" for n!, but there's this recursive definition:
0! = 1
(n + 1)! = (n + 1) x n!
2006-12-08 03:19:38 · answer #1 · answered by Anonymous · 0⤊ 0⤋
n! isproduct of all numbers starting from 1to n. Example:
6! = 1*2*3*4*5*6
2006-12-08 11:21:35 · answer #2 · answered by Manjari 2 · 0⤊ 0⤋
n!=n*(n-1)!, being 0!=1.
That's the formula. But for example, 4! would be:
4!=4*3*2*1=9.
For in-depth info, check http://en.wikipedia.org/wiki/Factorial
2006-12-08 11:23:45 · answer #3 · answered by Anonymous · 0⤊ 0⤋
n! = n * (n-1) * (n-2) * (n-3) * ... * 4 * 3 * 2 * 1
₢
2006-12-08 11:27:38 · answer #4 · answered by Luiz S 7 · 0⤊ 0⤋
N! ~ (2 Pi / (N+1))^1/2 E^-(N+1) (N+1)^(N+1)
For more information see the link below
2006-12-08 11:23:28 · answer #5 · answered by matty_boy1989 2 · 1⤊ 0⤋
n!=(n-1)*(n-2)*(n-3)*...*3*2*1*0!
where 0!=1
depends what n is...
if n=9 then
n!=9*8*7*6*5*4*3*2*1*0!
=9*8*7*6*5*4*3*2*1*1
2006-12-08 11:23:53 · answer #6 · answered by angel 2 · 0⤊ 0⤋
Those 'formulas' that people have posted are Sterling's approximation which is valid for large n. The only way to get n! is to grind it out.
2006-12-08 13:04:19 · answer #7 · answered by modulo_function 7 · 0⤊ 0⤋
What the others said. In spoken language, you say "n factorial". Simply leaping in the air and shouting "ENNNN!!!" doesn't cut it.
2006-12-08 11:27:15 · answer #8 · answered by Anonymous · 0⤊ 0⤋
0! = 1
n! = n * (n-1) * (n-2) * (n-3) * ... * 4 * 3 * 2 * 1 * 1
2006-12-08 11:22:51 · answer #9 · answered by Anonymous · 1⤊ 0⤋