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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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