Geoff is setting up an aquarium and must choose 4 of 6

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

by BTGmoderatorDC » Mon Apr 29, 2019 8:12 pm

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Geoff is setting up an aquarium and must choose 4 of 6 different fish and 2 of 3 different plants. How many different combinations of fish and plants can Geoff choose?

A. 8
B. 12
C. 18
D. 45
E. 90

OA D

Source: Princeton Review

Source: Beat The GMAT — Problem Solving | Mon Apr 29, 2019 8:12 pm

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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 » Mon Apr 29, 2019 9:13 pm

BTGmoderatorDC wrote:Geoff is setting up an aquarium and must choose 4 of 6 different fish and 2 of 3 different plants. How many different combinations of fish and plants can Geoff choose?

A. 8
B. 12
C. 18
D. 45
E. 90

OA D

Source: Princeton Review

Number of different combinations of fish and plants Geoff can choose = 6C4 * 3C2 = 6C2 * 3C1 = (6.5/1.2) * 3 = 45

The correct answer: D

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 » Thu May 02, 2019 4:21 pm

BTGmoderatorDC wrote:Geoff is setting up an aquarium and must choose 4 of 6 different fish and 2 of 3 different plants. How many different combinations of fish and plants can Geoff choose?

A. 8
B. 12
C. 18
D. 45
E. 90

OA D

Source: Princeton Review

The number of ways the fish can be selected is 6C4:

6! / (2! x (6 - 2)!) = 6!/(2! x 4!) = (6 x 5 x 4 x 3)/(4 x 3 x 2) = 15

The number of ways the plants can be selected is 3C2 = 3.

So the total number of different combinations is 15 x 3 = 45.

Answer: D

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