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# A city council will select 2 of 9 available firefighters and

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Senior Manager
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A city council will select 2 of 9 available firefighters and  [#permalink]

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27 May 2012, 09:08
5
3
00:00

Difficulty:

5% (low)

Question Stats:

82% (01:05) correct 18% (01:31) wrong based on 133 sessions

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A city council will select 2 of 9 available firefighters and 1 of 6 available police officers to serve on an advisory panel. How many different groups of 3 could serve on the panel?

A. 36
B. 72
C. 144
D. 216
E. 432
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Joined: 02 Sep 2009
Posts: 58320
Re: A city council will select 2 of 9 available firefighters and  [#permalink]

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28 May 2012, 02:04
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1
macjas wrote:
A city council will select 2 of 9 available firefighters and 1 of 6 available police officers to serve on an advisory panel. How many different groups of 3 could serve on the panel?

A. 36
B. 72
C. 144
D. 216
E. 432

Ways to select 2 firefighters out of 9 is $$C^2_9=36$$;
Ways to select 1 police officer out of 6 is $$C^1_6=6$$;

Total ways to select 2 firefighters out of 9 and 1 police officer out of 6 is 36*6=216.

It's called Principle of Multiplication: if one event can occur in m ways and a second can occur independently of the first in n ways, then the two events can occur in m*n ways.

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Re: A city council will select 2 of 9 available firefighters a  [#permalink]

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27 May 2012, 23:39
OA D

9C2 * 6C1

using the combination formula to choose 2 from 9 and 1 from 6
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Re: A city council will select 2 of 9 available firefighters and  [#permalink]

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08 Sep 2013, 00:09
The "how many different groups of 3" threw me at first, I thought I needed to exclude something, or say something like "we got 15 in total, so 5 groups of 3..." and so some calculations there.

Can someone please explain to me, in which case we would have divided the product of the two combinations by the total people selected out of the population? i.e. 15C5
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Joined: 02 Sep 2009
Posts: 58320
Re: A city council will select 2 of 9 available firefighters and  [#permalink]

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08 Sep 2013, 05:00
Skag55 wrote:
The "how many different groups of 3" threw me at first, I thought I needed to exclude something, or say something like "we got 15 in total, so 5 groups of 3..." and so some calculations there.

Can someone please explain to me, in which case we would have divided the product of the two combinations by the total people selected out of the population? i.e. 15C5

We would divide by 15C3 if the question were: what is the probability that when we select 3 people out of 9 firefighters and 6 police officers we get 2 firefighters and 1 police officer.
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Re: A city council will select 2 of 9 available firefighters and  [#permalink]

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08 Sep 2013, 10:30
Great, many thanks!
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Re: A city council will select 2 of 9 available firefighters and  [#permalink]

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11 Mar 2019, 08:15
macjas wrote:
A city council will select 2 of 9 available firefighters and 1 of 6 available police officers to serve on an advisory panel. How many different groups of 3 could serve on the panel?

A. 36
B. 72
C. 144
D. 216
E. 432

The number of ways to select the firefighters is 9C2 = 9!/(2! x 7!) = (9 x 8)/2! = 36.

The number of ways to select the police officers is 6C1 = 6s.

The total number of ways to select the group is 36 x 6 = 216.

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Re: A city council will select 2 of 9 available firefighters and   [#permalink] 11 Mar 2019, 08:15
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