Members of a student parliament took a vote on a proposition for a new social event on Fridays.

BTGModeratorVI wrote: ↑

Mon Apr 13, 2020 3:36 pm

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

Answer: D
Source: Veritas

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

BTGModeratorVI wrote: ↑

Mon Apr 13, 2020 3:36 pm

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

Answer: D
Source: Veritas

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. For more information about the FCP, watch this free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent