What is the formula for n!?

Answer 1

n! = n(n-1)(n-2)(n-3)...(3)(2)(1)

There is no actual formula to solve for n!, but there is an approximation formula, Stirling's approximation:

sqrt(2pi*n)*(n^n)*e^(-n).

It's much better than one percent accurate when calculating the log of the factorial of a large number.

Answer 2

Easiest way to think of it is, multiply the number by everything below it, down to 1:
2! = 2 x 1 = 2
3! = 3 x 2 x 1 = 6

And so on.

Answer 3

n! is just whatever n is times by all integers that are below it. for example, 6! would be 6x5x4x3x2x1.
if you have a calculator, then there should be a button for it. the highest number that works on a calculator, i think is 69. this is because the number just gets so big!
the formula for it would be:
nx(n-1)x(n-2)x...x((n-n)+1).
or nx(n-1)x(n-2)x...x1.
Is that any help?

Answer 4

When n gets too big for a calculator to show the answer, you can estimate it pretty accurately with "Stirling's Approximation". The simplest form of this is:

log(n!) ~= n * (log(n) - 1)

There are various more accurate forms, one of which is:

n! ~= sqrt(2 * PI * n) * n^n * exp(-n + 1/(12n))

Answer 5

I prefer the definition :

n! = 1*2*3*4*......*(n-1)*n

Answer 6

n! = n*(n-1)*(n-2)...(n-(n+1))

conditions:
n element of Natural numbers
0! = 1

examples:
2! = 2
3! = 6
4! = 24

Answer 7

n!=n(n-1)(n-2)(n-3)...(n-n+1)
with n being a naatural number

Answer 8

for n integers
0! = 1
n! = n*(n-1)!
this is product of all numbers from 1 to n.

Answer 9

this mite be a little advanced but:

n! = [integral from (x=0 --> infinity)] (x^(n-1)*e^(-x))

enjoy math

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Answer 10

1x2x3x....x(n-2)x(n-1)xn