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Ok say there are 10 teams and each team can win or lose..how many different outcomes can there be..i can't remember the formula for this can someone please help

2006-10-06 07:46:44 · 7 answers · asked by Anonymous in Science & Mathematics Mathematics

7 answers

If you mean that it is possible for all 10 teams to lose, it's like flipping 10 coins. Each has two possibilities so you would do 2x2x2x2x2x2x2x2x2x2.

2006-10-06 07:54:14 · answer #1 · answered by hayharbr 7 · 0 0

You don't mean that the teams have to play each other, right? Each team is separate? That means there are two possibilities for each team. The answer is to take the total amount of outcomes for team 1 (2) and multiply it my the same for team 2, all the way to the end: 2 to the 10th power = 2x2x2x2x2x2x2x2x2x2 = 1024 possible outcomes.

2006-10-06 14:52:41 · answer #2 · answered by comradivanred 2 · 0 0

Without any further restraints, any team has the chance to win or lose. That means that there are 10 possible outcomes. Team 1 wins, or team 2 wins, etc.
The possible outcomes are 10.

2006-10-06 22:25:44 · answer #3 · answered by Joseph G 3 · 0 0

If they're playing each other, then it's 10P2 (# of arrangements of 10 items taken 2 at a time where order is important)

10!/8!= 90 outcomes.

2006-10-06 15:12:37 · answer #4 · answered by Anonymous · 0 0

Could you please clarify this question? Are the ten teams playing each other, for a total of 5 games? Or what?

2006-10-06 14:58:34 · answer #5 · answered by James L 5 · 0 0

This looks like a binomial problem

2006-10-06 15:11:06 · answer #6 · answered by Bill N 3 · 0 0

10 factorial! = 3628800!

2006-10-06 14:52:50 · answer #7 · answered by safety first 3 · 0 0

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