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A committee has 10 members. How many different selections o

by Max@Math Revolution » Tue Feb 13, 2018 11:57 pm
[GMAT math practice question]

A committee has 10 members. How many different selections of a chairman, an editor, and a secretary can be made from the members of the committee?

A. 180
B. 360
C. 540
D. 720
E. 810
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by elias.latour.apex » Wed Feb 14, 2018 5:45 am
Imagine that we are going to choose the chairman first. We have 10 different people we can choose from. Then we will choose the editor. Assuming that the same person cannot hold both positions (the problem doesn't say) that means that we will only have 9 people to choose from. Similarly, when we choose the secretary, we will have 8 possibilities.

Accordingly, the total number of possibilities is 10*9*8 = 720.
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by Scott@TargetTestPrep » Thu Feb 15, 2018 10:18 am
Max@Math Revolution wrote:[GMAT math practice question]

A committee has 10 members. How many different selections of a chairman, an editor, and a secretary can be made from the members of the committee?

A. 180
B. 360
C. 540
D. 720
E. 810
The order of selection is important in this scenario, so we use the permutation formula. The number of ways to select a chairman, an editor, and a secretary from 10 people is 10P3 = 10!/(10 - 3)! = 10!/7! = 10 x 9 x 8 = 720.

Answer: D

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by Max@Math Revolution » Fri Feb 16, 2018 3:57 am
=>

There are 10 ways of choosing the chairman. Once the chairman has been chosen, there are 9 remaining possibilities for the editor. Once both the chairman and the editor have been chosen, 8 choices remain for the secretary. So, the total number of choices for these three positions is:
10*9*8 = 720.
This may also be written as
10P3 = 10 * 9 * 8 = 720.

Therefore, the answer is D.

Answer: D
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