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Probability question

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Probability question

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If 3 persons are selected at random from 8 persons for 3 positions present, vice-present, and secretary, how many such ways are possible?
A. 56
B. 70
C. 240
D. 336
E. 1680
The QA to this question is d.
Can you show your workings to this question and also, why is B not correct because it's also close.

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Hi Roland2rule,

Since the roles of President, Vice-President and Secretary are distinct, this is actually a Permutation question (and not a Probability question). With 8 possible people to fill the 3 roles - and the restriction that 3 different people must be chosen, we have:

8 possible choices for President. Once one person is chosen, we have...
7 possible choices for Vice-President. Once one person is chosen, we have...
6 possible choices for Secretary.
(8)(7)(6) = 336 possible options

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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GMAT/MBA Expert

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Roland2rule wrote:
If 3 persons are selected at random from 8 persons for 3 positions present, vice-present, and secretary, how many such ways are possible?
A. 56
B. 70
C. 240
D. 336
E. 1680
Take the task of filling the 3 positions and break it into stages.

Stage 1: Select someone to be president
Since there are 8 people to choose from, we can complete stage 1 in 8 ways

Stage 2: Select someone to be vice president
There are 7 people remaining to choose from (since we already selected a person in stage 1), so we can complete stage 2 in 7 ways.

Stage 3: Select someone to be secretary
There are 6 people remaining to choose from, so we can complete stage 3 in 6 ways.

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus fill the 3 positions ) in (8)(7)(6) ways (= 336 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 our free video: http://www.gmatprepnow.com/module/gmat-counting/video/775

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

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- http://www.beatthegmat.com/what-should-be-the-answer-t267256.html
- http://www.beatthegmat.com/counting-problem-company-recruitment-t244302.html
- http://www.beatthegmat.com/picking-a-5-digit-code-with-an-odd-middle-digit-t273110.html
- http://www.beatthegmat.com/permutation-combination-simple-one-t257412.html
- http://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- http://www.beatthegmat.com/combinatorics-solution-explanation-t273194.html
- http://www.beatthegmat.com/arabian-horses-good-one-t150703.html
- http://www.beatthegmat.com/sub-sets-probability-t273337.html
- http://www.beatthegmat.com/combinatorics-problem-t273180.html
- http://www.beatthegmat.com/digits-numbers-t270127.html
- http://www.beatthegmat.com/doubt-on-separator-method-t271047.html
- http://www.beatthegmat.com/combinatorics-problem-t267079.html


DIFFICULT
- http://www.beatthegmat.com/wonderful-p-c-ques-t271001.html
- http://www.beatthegmat.com/permutation-and-combination-t273915.html
- http://www.beatthegmat.com/permutation-t122873.html
- http://www.beatthegmat.com/no-two-ladies-sit-together-t275661.html
- http://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent

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Roland2rule wrote:
If 3 persons are selected at random from 8 persons for 3 positions present, vice-present, and secretary, how many such ways are possible?
A. 56
B. 70
C. 240
D. 336
E. 1680
The order of selection is important in this scenario, so this is a permutation problem. The number of ways to select 3 people from 8 for the given positions is 8P3 = 8!/(8-3)! = 8 x 7 x 6 = 336 ways.

Answer: D

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Scott Woodbury-Stewart
Founder and CEO
scott@targettestprep.com



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