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

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Math Expert
Joined: 02 Sep 2009
Posts: 58381
Geoff is setting up an aquarium and must choose 4 of 6 different fish  [#permalink]

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09 Jul 2018, 05:19
00:00

Difficulty:

5% (low)

Question Stats:

79% (00:53) correct 21% (01:04) wrong based on 72 sessions

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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

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Geoff is setting up an aquarium and must choose 4 of 6 different fish  [#permalink]

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09 Jul 2018, 05:33
Number of ways of selecting 4 fish from 6 fish are $$6c_4$$ = $$\frac{6!}{4!2!}$$ = 15

Number of ways of selecting 2 plants from 3 plants are $$3c_2$$ = $$\frac{3!}{2!}$$ = 3

Number of different combinations of fish and plants Geoff can choose = 15 * 3 = 45

Hence option D
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Re: Geoff is setting up an aquarium and must choose 4 of 6 different fish  [#permalink]

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09 Jul 2018, 05:42
Bunuel 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

No of combinations of fish and plant = no of ways of choosing 4 dishes out of 6* no of ways of choosing 2 plants out of 3=6C4*3C2=5*3=15

Ans. (D)

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Re: Geoff is setting up an aquarium and must choose 4 of 6 different fish  [#permalink]

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09 Jul 2018, 05:44

Solution

Given:
• Geoff is setting up an aquarium and must choose 4 of 6 different fish and 2 of 3 different plants

To find:
• In how many different combinations Geoff can choose fish and plants

Approach and Working:
Geoff needs to choose 4 out of 6 different fish
• Number of ways he can choose fish = $$^6C_4$$ = 15

Geoff needs to choose 2 out of 3 different plants
• Number of ways he can choose plant = $$^3C_2$$ = 3

Therefore, total number of ways Geoff can make the choices = 15 * 3 = 45

Hence, the correct answer is option D.

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Re: Geoff is setting up an aquarium and must choose 4 of 6 different fish  [#permalink]

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09 Jul 2018, 06:33
+1 for D.

Number of ways to choose fish = 6C4
Number of ways to choose plant = 3C2

Total number of ways to make the choices = 6C4*3C2 = 15 * 3 = 45

Hence, D.
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Re: Geoff is setting up an aquarium and must choose 4 of 6 different fish  [#permalink]

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11 Jul 2018, 01:42
Bunuel 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

this is an AND question
4 fish selected out of 6, (sequence of selection doesn't matter ), total combinations =$$\frac{6!}{4!*2!}$$

2 plants selected out of 3, (sequence of selection doesn't matter ), total combinations =$$\frac{3!}{1!*2!}$$

=$$\frac{6!}{4!*2!}$$ * $$\frac{3!}{1!*2!}$$ = 45
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Re: Geoff is setting up an aquarium and must choose 4 of 6 different fish  [#permalink]

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08 Mar 2019, 08:04
Bunuel 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

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 different fish   [#permalink] 08 Mar 2019, 08:04
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