Seven basketball teams play in a league against each other. At the end of the season, how many different arrangements are there for the top 3 teams in the rankings?
A. 6
B. 42
C. 210
D. 5,040
E. 50,450
The OA is C.
I'm really confused by this PS question. Experts, any suggestion about how to solve it? Thanks in advance.
Seven basketball teams play in a league against each other
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Take the task of arranging the top 3 teams and break it into stages.LUANDATO wrote:Seven basketball teams play in a league against each other. At the end of the season, how many different arrangements are there for the top 3 teams in the rankings?
A. 6
B. 42
C. 210
D. 5,040
E. 50,450
Stage 1: Select the 1st place team
There are 7 teams to choose from, so we can complete stage 1 in 7 ways
Stage 2: Select the 2nd place team
Since already selected a team in stage 1, there are now 6 teams remaining to choose from.
So, we can complete stage 2 in 6 ways
Stage 3: Select the 3rd place team
There are now 5 teams remaining to choose from.
So, we can complete stage 3 in 5 ways
By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus arrange the top 3 teams) in (7)(6)(5) ways (= 210 ways)
Answer: C
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Since the order matters, the number of ways to choose 3 teams from 7 and arrange them is:BTGmoderatorLU wrote:Seven basketball teams play in a league against each other. At the end of the season, how many different arrangements are there for the top 3 teams in the rankings?
A. 6
B. 42
C. 210
D. 5,040
E. 50,450
7P3 = 7!/(7-3)! = 7 x 6 x 5 = 210
Answer: C
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