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

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Geoff is setting up an aquarium and must choose 4 of 6 diffe [#permalink]  30 Dec 2017, 13:31
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Question Stats:

100% (00:35) correct 0% (00:00) wrong based on 3 sessions
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

Drill 1
Question: 7
Page: 560

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Sandy
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GRE Prep Club Legend
Joined: 07 Jun 2014
Posts: 4809
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 123

Kudos [?]: 1983 [0], given: 397

Re: Geoff is setting up an aquarium and must choose 4 of 6 diffe [#permalink]  07 Jan 2018, 18:16
Expert's post
Explanation

First find the number of groups of fish he can select. This is your number of slots: _ _ _ __.

There are six fish he can choose for the first slot, 5 for the second and so on: 6543.

Since order doesn’t matter, you need to divide by the factorial of the number of slots: $$\frac{6213}{4321}$$. Reduce your number to get 3 × 5 = 15.

For plants you have two slots so = $$3/2 \times 2/1 = 3.3 \times 15 =45$$.

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Sandy
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Re: Geoff is setting up an aquarium and must choose 4 of 6 diffe [#permalink]  08 Mar 2019, 07:04
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sandy 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

Drill 1
Question: 7
Page: 560

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.

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Re: Geoff is setting up an aquarium and must choose 4 of 6 diffe [#permalink]  09 Mar 2019, 02:11
Expert's post
Thanks for the other approach Sir.

Regards
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Re: Geoff is setting up an aquarium and must choose 4 of 6 diffe   [#permalink] 09 Mar 2019, 02:11
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