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A committee has 10 members. How many different selections o

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A committee has 10 members. How many different selections o

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[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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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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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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scott@targettestprep.com



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=>

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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Only $149 for 3 month Online Course
Free Resources-30 day online access & Diagnostic Test
Unlimited Access to over 120 free video lessons-try it yourself
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