Bunuel
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.
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Fundamental Counting Principle (FCP)
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