Probability question

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

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
--------------------------

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: 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

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