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Re: A city council will select 2 of 9 available firefighters and [#permalink]
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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Re: A city council will select 2 of 9 available firefighters and [#permalink]
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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]
Great, many thanks!
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Re: A city council will select 2 of 9 available firefighters and [#permalink]
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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.

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