The final score of a soccer game is 6 to 3. How many different scores were possible at half time? Permutations and Organized Counting Question
The winning team had 7 possible scores at half time (0, 1, 2, 3, 4, 5, 6) and the losing team had 4 possible scores at half time (0, 1, 2, 3) I think it's safe to just multiply the number of possible scores each team had together to get the total number of combinations. So 28?
yea the answer is 28, thanks man
oh wait dude can you explain why u have to multiply the the possibilities?
Think of it like this: If the winners had 0 at half time, the losers could have had 0, 1, 2 or 3 That's 4 possible scores If the winners had 1 at half time, the losers could have had 0, 1, 2 or 3 That's another 4 ... If you keep going and adding up the possible combinations, it's the same as 7 times 4 You don't need to discount any "duplicates" (like 1-0 and 0-1) because they are different scores (a different team is winning). The scores of each team are independent (that wouldn't be the case if, for example, the question specified that one team was in the lead at half time).
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