Five people are running in a race. The first three to finish

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Thu Aug 08, 2019 2:34 am

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Five people are running in a race. The first three to finish win gift certificates. How many different groups of people could win the gift certificates?

A. 5
B. 10
C. 60
D. 120
E. 125

OA B

Source: Princeton Review

Source: Beat The GMAT — Problem Solving | Thu Aug 08, 2019 2:34 am

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 amFollowed by: 34 members; Location: Grand Central / New York; Thanked: 470 times

by Jay@ManhattanReview » Fri Aug 09, 2019 12:58 am

BTGmoderatorDC wrote:Five people are running in a race. The first three to finish win gift certificates. How many different groups of people could win the gift certificates?

A. 5
B. 10
C. 60
D. 120
E. 125

OA B

Source: Princeton Review

Since any 3 out of 5 can win gift certificates, the number of different groups of people could win the gift certificates = 5C3 = 5C2 = (5.4) / (1.2) = 10 groups

Though the order of first, second and third places would matter, it does not matter in winning gift certificates. For example, D being first, B being second and A being third AND B being first, A being second and D being third is a group with the same 3 people.

The correct answer: B

Hope this helps!

-Jay
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Scott@TargetTestPrep: GMAT Instructor; Posts: 8246; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Sun Aug 11, 2019 6:29 pm

BTGmoderatorDC wrote:Five people are running in a race. The first three to finish win gift certificates. How many different groups of people could win the gift certificates?

A. 5
B. 10
C. 60
D. 120
E. 125

OA B

Source: Princeton Review

The number of groups of people who could win a gift certificate is 5C3 = (5 x 4 x 3)/3! = 10.

Answer: B

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