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

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BTGmoderatorLU: Moderator; Posts: 2505; Joined: Sun Oct 15, 2017 1:50 pm; Followed by:6 members

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

by BTGmoderatorLU » Tue Mar 05, 2019 2:41 pm

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Source: Princeton Review

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

The OA is D

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Source: Beat The GMAT — Problem Solving | Tue Mar 05, 2019 2:41 pm

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

by Jay@ManhattanReview » Wed Mar 06, 2019 2:34 am

BTGmoderatorLU wrote:Source: Princeton Review

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

The OA is D

Number of ways of choosing different fish out of 6 = 6C4 = 6C2 = 6.5 / 1.2 = 15;

Number of ways of choosing 2 different plants out of 3 = 3C2 = 3C1 = 3

Total number of ways of setting up an aquarium = 15*3 = 45

The correct answer: D

Hope this helps!

-Jay
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Scott@TargetTestPrep: GMAT Instructor; Posts: 8083; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Fri Mar 08, 2019 6:48 am

BTGmoderatorLU wrote:Source: Princeton Review

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

The OA is D

The number of fish can be selected in 6C4 ways:

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

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

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

Answer: D

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