COMBINATION Problem

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

COMBINATION Problem

by BTGmoderatorRO » Sun Sep 10, 2017 2:34 pm

What is the number of different ways to choose a chairman, two deputies, and two assistants for the class committee out of 7 students up for elections.

Which solution is correct ? And Why ?

A) (7C1) * (6C1) * (5C1) * (4C1)* (3C1)
B) (7C1) * (6C2) * (4C3)
C) (7C1) * (6C2) * (4C2)
D) (7C1) * (6C1) * (5C1) * (4C1*4C1)* (3C1*3C2)
E) (7C1*7C2) * (6C2) * (4C2)

The most accurate and correct option is Option C

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Source: Beat The GMAT — Problem Solving | Sun Sep 10, 2017 2:34 pm

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by [email protected] » Sun Sep 10, 2017 3:11 pm

Hi Roland2rule,

Based on the wording of the prompt, we're meant to assume that the roles of 'deputy' and 'assistant' are NOT unique - meaning that if A and B are the two deputies, then that would be considered the SAME option as having B and A as the two deputies (and that the same logic would apply for the two assistants). Thus, we can work through the 'list' of jobs one calculation at a time.

-First, we have 7 students and 1 chairman --> that's 7c1 (or simply put: 7).
-After we select the chairman, we have 6 students left for the 2 deputies --> that's 6c2
-Finally, after we select the deputies, there are 4 students left for the 2 assistants --> that's 4c2

Total Number of possible groups: (7)(6c2)(4c2)

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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Re: COMBINATION Problem

by Scott@TargetTestPrep » Sun Feb 09, 2020 5:34 am

BTGmoderatorRO wrote: ↑
Sun Sep 10, 2017 2:34 pm
What is the number of different ways to choose a chairman, two deputies, and two assistants for the class committee out of 7 students up for elections.

Which solution is correct ? And Why ?

A) (7C1) * (6C1) * (5C1) * (4C1)* (3C1)
B) (7C1) * (6C2) * (4C3)
C) (7C1) * (6C2) * (4C2)
D) (7C1) * (6C1) * (5C1) * (4C1*4C1)* (3C1*3C2)
E) (7C1*7C2) * (6C2) * (4C2)

The most accurate and correct option is Option C

There are 7 options for the chairman, 6 options for the first deputy, 5 options for the second deputy, 4 options for the first assistant and 3 options for the second assistant. Since the two deputy positions and two assistant positions are indistinguishable, the total number of ways to choose the committee is (7 x 6 x 5 x 4 x 3)/(2 x 2).

Notice that 6C2 = 6!/(4!*2!) = (6 x 5)/2 and 4C2 = 4!/(2!*2!) = (4 x 3)/2. Thus, the number of ways to select one chairman two deputies, and two assistants is:

7C1 x 6C2 x 4C2

Answer: C

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