If eleven basketball teams are in a tournament, find the number of different ways that first, second, and third place can be decided, assuming ties are not allowed.
A)17,160
B)6
C)1,320
D)11,880
E)132

Question

Quizplus · Accepted Answer

This problem can be solved using permutations since the order in which the teams are placed matters. There are 11 teams to choose from for the first place, 10 teams left for the second place, and 9 teams left for the third place. Therefore, the number of different ways to decide first, second, and third places is 11 * 10 * 9 = 990. However, the correct calculation should be 11 * 10 * 9 = 990, indicating a mistake in my initial explanation. The correct approach to calculate the number of ways to decide first, second, and third places among 11 teams, considering the order matters (without repetition), is indeed using permutations. The correct formula for this scenario is P(n, r) = n! / (n-r)!, where n is the total number of teams and r is the number of places to be filled. Here, n = 11 (total teams) and r = 3 (places to be filled - first, second, and third). Thus, the correct calculation is P(11, 3) = 11! / (11-3)! = 11! / 8! = 11 * 10 * 9 = 990. This indicates an error in my response regarding the correct answer and explanation. The correct answer should reflect the calculation based on permutations for 11 teams and 3 places.

If Eleven Basketball Teams Are in a Tournament, Find the Number