How many outcomes could their be, not including ties, or a push. Just win or lose.
Thanks!
2006-09-24 18:25:25 · 9 answers · asked by maxmelton@sbcglobal.net 1 in Sports ➔ Football (American)
1) 6 - 6
2) 8 - 4
3) 2 - 10
4) 10 - 2
5) 4 - 8
6) 1 - 11
7) 11 - 1
8) 0 - 12
9) 12 - 0
7) 7 - 5
8) 5 - 7
9) 3 - 9
10) 9 - 3
2006-09-24 18:30:50 · answer #1 · answered by ckpuppybear2002 4 · 0⤊ 0⤋
well you can only have 12 different final records, but to get there, you calculate each game with 2 possible outcomes, and you do that 12 times, so 2^12=4096 different potential won/loss schedules for a team
2006-09-25 02:22:16 · answer #2 · answered by C_Millionaire 5 · 0⤊ 0⤋
try use this formula... it works for most of the numbers...
N(N-1) divide by 2 Whereas N is the Number....
Eg. in this case is 12... So 12 x (12-1) divide by 2 = 66
So combination should be 66
2006-09-25 04:04:11 · answer #3 · answered by porridge0000 3 · 0⤊ 0⤋
144
2006-09-25 02:35:44 · answer #4 · answered by pato 2 · 0⤊ 0⤋
Well I do'nt know what a push is but the answer to your question is 13.
2006-09-25 05:06:32 · answer #5 · answered by Lance D 1 · 0⤊ 0⤋
Don't let keoni get to you. That's his standard response for every question he answered ('cept you weren't called lazy and I was! lol) He's just an @sshole.
2006-09-25 14:53:09 · answer #6 · answered by Sarah 2 · 0⤊ 0⤋
try the Mathematics category, nerd
2006-09-27 02:09:41 · answer #7 · answered by Anonymous · 0⤊ 0⤋
uhhh do the math on a piece of paper?
2006-09-25 01:28:37 · answer #8 · answered by keoni_21 3 · 0⤊ 0⤋
2x2x2x2x2x2x2x2x2x2x2x2=4096
2006-09-25 01:32:06 · answer #9 · answered by bearhill13 2 · 0⤊ 0⤋