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Members of a student parliament took a vote on a proposition for a new

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New post 23 Feb 2017, 12:15
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Members of a student parliament took a vote on a proposition for a new social event on Fridays. If all the members of the parliament voted either for or against the proposition, and if the proposition was accepted in a 5-to-2 vote, how many ways could the members have voted?

A. 7
B. 10
C. 14
D. 21
E. 42

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Re: Members of a student parliament took a vote on a proposition for a new  [#permalink]

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New post 23 Feb 2017, 12:36
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Bunuel wrote:
Members of a student parliament took a vote on a proposition for a new social event on Fridays. If all the members of the parliament voted either for or against the proposition, and if the proposition was accepted in a 5-to-2 vote, how many ways could the members have voted?

A. 7
B. 10
C. 14
D. 21
E. 42


The 5-to-2 vote tells us that there are 7 members of a student parliament.
It also tells us that 5 students voted FOR the proposition, and 2 students voted AGAINST the proposition.

So, in how many ways can we choose 5 of the 7 students to be the ones who voted FOR the proposition.
Once we select the 7 students to be the ones who voted FOR the proposition, we have also indirectly selected the 2 students who voted AGAINST the proposition. Those 2 students will be the ones who weren't selected to be the ones who voted FOR the proposition.

So, let's take the task dividing the students into 2 groups (those FOR the proposition and those AGAINST the proposition and break it into stages.

Stage 1: Select 5 students to be the ones those FOR the proposition
Since the order in which we select the 5 students does not matter, we can use combinations.
We can select 5 students from 7 students in 7C5 ways (21 ways)
So, we can complete stage 1 in 21 ways

Stage 2: Select 2 students to be the ones those AGAINST the proposition
Once we complete stage 1, there only 2 students remaining.
By default, those 2 students are AUTOMATICALLY the ones AGAINST the proposition.
So we can complete this stage in 1 way (both are in the AGAINST camp)

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus arrange all 7 students) in (21)(1) ways (= 21 ways)

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT, so be sure to learn this technique.

RELATED VIDEOS FROM OUR COURSE
Fundamental Counting Principle (FCP)



Fundamental Counting Principle - example



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Re: Members of a student parliament took a vote on a proposition for a new  [#permalink]

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New post 24 Feb 2017, 06:09
Bunuel wrote:
Members of a student parliament took a vote on a proposition for a new social event on Fridays. If all the members of the parliament voted either for or against the proposition, and if the proposition was accepted in a 5-to-2 vote, how many ways could the members have voted?

A. 7
B. 10
C. 14
D. 21
E. 42


5 to 2 vote i.e. we need 5 out of 7 in favour ...

5 out of 7 can be chosen in a total of 7C5 ways and 7C5=21

Answer: OPtion D
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Re: Members of a student parliament took a vote on a proposition for a new  [#permalink]

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New post 27 Feb 2017, 11:58
Bunuel wrote:
Members of a student parliament took a vote on a proposition for a new social event on Fridays. If all the members of the parliament voted either for or against the proposition, and if the proposition was accepted in a 5-to-2 vote, how many ways could the members have voted?

A. 7
B. 10
C. 14
D. 21
E. 42


Since the proposition was accepted in a 5-to-2 vote, that means 5 members say “yes” and the other 2 members say “no.” Therefore, the problem becomes: In how many ways can we arrange 5 yesses and 2 nos? That is, how many ways can we arrange the Ys and Ns in the string YYYYYNN? The answer is:

7!/(5! x 2!) = (7 x 6 x 5!)/(5! x 2 x 1) = 42/2 = 21

Answer: D
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Re: Members of a student parliament took a vote on a proposition for a new   [#permalink] 27 Feb 2017, 11:58
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