Re: [OT] Math

> Nathan wrote:
> Again, here we should have:
> 
> P = 6C4*(1/6)^4*(5/6)^2 + 6C5*(1/6)^5*(5/6) + 6C6*(1/6)^6 = 0.87% =
0.9%
> (again as you got).
> 
> In general, when you want to know whether or not exactly n events,
with
> P=a will occur out of a set of m events, m>n,
> 
> P = mCn*(a)^n*(1-a)^(m-n)
> 
> Hope that helps,
>	-Brian Quirt
FYI,

I didn't see anything about how to calculate combinatorials (sp?). 
Maybe
I just missed it, so here goes:

mCn = (m)! / (n! (m-n)!)

So in the above case you've got 6C4 = (6!) / (4! (6-4)!)
= 6 * 5 * 4 * 3 * 2 * 1 / 4 * 3 * 2 * 2 * 1 * 1
= 15
There are 15 ways that 4 items can be arranged inside of 6 items.

Hope that helps,