2006-11-01 07:05:12 · 6 answers · asked by Sonja - 1 in Science & Mathematics ➔ Mathematics
2^8 = 256 ways.
This again assumes that order doesn't matter.
Include or exclude person A = 2 ways
Include or exclude person B = 2 ways
...
Include or exclude person H = 2 ways.
So the answer is 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2^8 = 256
2006-11-01 07:13:00 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
if it is combinations of two, without doubles, then simply think about it. If A can go with B, C, D, etc, then there are 7 combinations for A. Since B already went with A, he can now only go with C, D, E, etc for 6 combinations. C already went with A and B, so only D, E, F, etc, for 5 combinations.
so, 7+6+5+4+...down to 1.
If the order of matters, then you have 8 choices that can be person #1, and 7 choices that can be person #2, so a total of 56 combinations where the order of the person matters.
2006-11-01 15:11:54 · answer #2 · answered by romanwahoo 2 · 0⤊ 0⤋
Depends on what type of combinations you're talking about. For example, if you have eight people and you want to know how many boy/girl combinations there could be, that would be 2 to the eighth power, or 256.
2006-11-01 15:12:33 · answer #3 · answered by sarge927 7 · 0⤊ 0⤋
64
2006-11-01 15:06:30 · answer #4 · answered by jon 4 · 0⤊ 0⤋
8!
2006-11-01 15:12:07 · answer #5 · answered by Anonymous · 0⤊ 0⤋
8!=8*7*6*5*4*3*2*1=40320 (WOW!!)
2006-11-01 15:14:12 · answer #6 · answered by Little Fairy 4 · 0⤊ 0⤋