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by BTGmoderatorDC » Mon Oct 29, 2018 4:15 am

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Employees at a company will vote for an executive team of five people from eight qualified candidates. The executive team consists of a president, a treasurer, and three warrant officers. If an executive team is considered different if any of the same people hold different offices, then how many possible executive teams could be selected from the eight candidates?

A. 56
B. 120
C. 210
D. 1120
E. 6720

OA D

Source: Magoosh
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by GMATGuruNY » Mon Oct 29, 2018 4:35 am
BTGmoderatorDC wrote:Employees at a company will vote for an executive team of five people from eight qualified candidates. The executive team consists of a president, a treasurer, and three warrant officers. If an executive team is considered different if any of the same people hold different offices, then how many possible executive teams could be selected from the eight candidates?

A. 56
B. 120
C. 210
D. 1120
E. 6720
Number of options for president = 8. (Any of the 8 candidates.)
Number of options for treasurer = 7. (Any of the 7 remaining candidates.)
From the 6 remaining people, the number of ways to choose 3 to serve as warrant officers = 6C3 = (6*5*4)/(3*2*1) = 20.
To combine these options, we multiply:
8*7*20 = 1120.

The correct answer is D.
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by fskilnik@GMATH » Mon Oct 29, 2018 7:29 am
BTGmoderatorDC wrote:Employees at a company will vote for an executive team of five people from eight qualified candidates. The executive team consists of a president, a treasurer, and three warrant officers. If an executive team is considered different if any of the same people hold different offices, then how many possible executive teams could be selected from the eight candidates?

A. 56
B. 120
C. 210
D. 1120
E. 6720
Source: Magoosh
\[? = \underbrace {C\left( {8,1} \right)}_{{\text{president}}} \cdot \underbrace {C\left( {7,1} \right)}_{{\text{treasurer}}} \cdot \underbrace {C\left( {6,3} \right)}_{{\text{3}}\,\,{\text{officers}}} = 8 \cdot 7 \cdot \frac{{6 \cdot 5 \cdot 4}}{{3 \cdot 2}} = 56 \cdot 20 = 1220\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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Portuguese-speakers :: https://www.gmath.com.br
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by Scott@TargetTestPrep » Tue Oct 30, 2018 6:09 pm
BTGmoderatorDC wrote:Employees at a company will vote for an executive team of five people from eight qualified candidates. The executive team consists of a president, a treasurer, and three warrant officers. If an executive team is considered different if any of the same people hold different offices, then how many possible executive teams could be selected from the eight candidates?

A. 56
B. 120
C. 210
D. 1120
E. 6720
If each of the 5-member executive team has a different position, then there can be 8P5 ways to choose the 5 people from a total of 8 people. However, since 3 of 5-member team have the same position (warrant officer) and it doesn't matter how we arrange them, we have to divide 8P5 by 3!. So the answer is:

8P5/3! = (8 x 7 x 6 x 5 x 4)/(3 x 2) = 8 x 7 x 5 x 4 = 1120

Alternate Solution:

There are 8 candidates for president, and after the president is chosen, there are 7 remaining candidates for the position of treasurer. After these positions are filled, the remaining 3 positions can be filled by any of the 6 remaining members, therefore there are 6C3 = 6!/(3!*3!) = 20 ways of such a choice. In total, there are 8 x 7 x 20 = 1120 ways of forming the executive team.

Answer: D

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