How many different outcomes can there be from 9 football games combined? What is the equation to figure this?

Question

Is it 2 to the 9th power?

Answer 1

3 to the 9th power, they can tie as well.

Answer 2

It depends on what you consider a different outcome.

Let teams A and B play against each other, and let team A either win or lose.

If A wins the first game, then loses the rest, do we consider this a different outcome than A losing the first eight games, then winning the ninth? If so, the number of different outcomes is 2^9.

If we do not consider those to be different events, and we're only interested in the number of games won by team A or B, then there are only 10 different outcomes, corresponding to team A winning 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 games.

Answer 3

If there are only two teams then, yes, the answer is 2^9.

If there are three teams then it's 3^9, etc..

I don't understand what you mean by an equation to figure this.. I just look at it as a counting problem counting in base 2 (or three or however many teams there are.)

For example, say the teams are 0 and 1. Here is how I look at how many combinations there are: (like counting in base two with 9 bits)

000000000
000000001
000000010
000000011
000000100
000000101
000000110
000000111
000001000
000001001
000001010
000001011
000001100
000001101
000001110
000001111
000010000
...
111111100
111111101
111111110
111111111

if there were three teams it would look like this:

000000000
000000001
000000002
000000010
000000011
000000012
000000020
000000021
000000022
000000100
000000101
...
222222222

Answer 4

Assuming that each game has one winner and one loser, your answer is correct.

There are two possible outcomes for each game, and 9 independent games, so the number of outcomes is

2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 2^9 = 512.