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Seven basketball teams play in a league against each other. At the end of the season, how many different arrangements

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Seven basketball teams play in a league against each other. At the end of the season, how many different arrangements

by BTGmoderatorDC » Sun Dec 20, 2020 6:02 pm

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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


OA C

Source: Princeton Review
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Re: Seven basketball teams play in a league against each other. At the end of the season, how many different arrangement

by Brent@GMATPrepNow » Mon Dec 21, 2020 6:00 am
BTGmoderatorDC wrote: ↑
Sun Dec 20, 2020 6:02 pm
 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


OA C

Source: Princeton Review
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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Re: Seven basketball teams play in a league against each other. At the end of the season, how many different arrangement

by swerve » Mon Dec 21, 2020 10:58 am
BTGmoderatorDC wrote: ↑
Sun Dec 20, 2020 6:02 pm
 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


OA C

Source: Princeton Review
It's simply 7P3 = 7!/(7-3)! = 210. Selecting 3 out of 7 when the order of the selection matters.

Therefore, C
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