Dear smart people... (not GalCiv2 related)

For the first I think its 6^3 combinations..not sure been a while since I did any of that kind of thing.

For the next sorry, I don't remember, I'd look it up in one of my math books, but I don't want to wake up my girlfriend.

Since I'm pretty sure you're trying to figure out the bell curve distribution for a roll of three dices, I'll ignore the implication of unique rolls and all that good stuff to focus on the numbers you're looking for...

Basically, rolling 3 6-sided dices gives you 6^3 = 216 combinations...

Then, to figure out the number of combinations, you need to figure out how many combination of 3 numbers add up to that number. Normally there are formulas for this, but I'm afraid people tend to get brain fried looking at them, so let me put it down in terms people can understand...

3 = 1 (1+1+1)
4 = 3 (2+1+1), (1+2+1), (1+1+2) (remember, it's not unique)
5 = 6 (3+1+1)x3, (2+2+1)x3 (2 different # produces 3 combinations
6 = 10 (4+1+1)x3, (3+2+1)x6, (2+2+2) (3 different # produces 6 combinations)
7 = 15 (5+1+1)x3, (4+2+1)x6, (3+3+1)x3, (3+2+2)x3
8 = 21 (6+1+1)x3, (5+2+1)x6, (4+3+1)x6, (4+2+2)x3, (3+3+2)x3
9 = 25 (6+2+1)x6, (5+3+1)x6, (5+2+2)x3, (4+4+1)x3, (4+3+2)x6, (3+3+3)
10 = 27 (6+3+1)x6, (6+2+2)x3, (5+4+1)x6, (5+3+2)x6, (4+4+2)x3, (4+3+3)x3
11 = 27 (6+4+1)x6, (6+3+2)x6, (5+5+1)x3, (5+4+2)x6, (5+3+3)x3, (4+4+3)x3
12 = 25 (6+5+1)x6, (6+4+2)x6, (6+3+3)x3, (5+5+2)x3, (5+4+3)x6, (4+4+4)
13 = 21 (6+6+1)x3, (6+5+2)x6, (6+4+3)x6, (5+5+3)x3, (5+4+4)x3
14 = 15 (6+6+2)x3, (6+5+3)x6, (6+4+4)x3, (5+5+4)x3
15 = 10 (6+6+3)x3, (6+5+4)x6, (5+5+5)
16 = 6 (6+6+4)x3, (6+5+5)x3
17 = 3 (6+6+5)x3
18 = 1 (6+6+6)

To check... (1+3+6+10+15+21+25+27) = 108 x 2 = 216 total combinations.

I'm sure you can take care of the rest...

The answer is 42.

nice.

[moved to offtopic]

Thanks you guys, and thanks Kyro for moving it, hehe I meant to put it there in the first place...

Cool this was exactly what I wanted to know, you all are awesome!