Seven basketball teams play in a league against each other

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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.

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by Brent@GMATPrepNow » Wed Mar 14, 2018 3:17 pm
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
Take the task of arranging the top 3 teams and break it into stages.

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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Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Sun Jun 02, 2019 4:22 pm
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
Since the order matters, the number of ways to choose 3 teams from 7 and arrange them is:

7P3 = 7!/(7-3)! = 7 x 6 x 5 = 210

Answer: C

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