Probability question

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BTGmoderatorRO: Moderator; Posts: 772; Joined: Wed Aug 30, 2017 6:29 pm; Followed by:6 members

Probability question

by BTGmoderatorRO » Sat Sep 30, 2017 7:18 pm

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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Source: Beat The GMAT — Problem Solving | Sat Sep 30, 2017 7:18 pm

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by [email protected] » Sun Oct 01, 2017 10:42 am

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

Contact Rich at [email protected]

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by Brent@GMATPrepNow » Mon Oct 02, 2017 7:59 am

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 ([spoiler]= 336 ways[/spoiler])

Answer: D
--------------------------

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8085; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Thu Jan 04, 2018 9:25 am

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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