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Difficulty:
75%
(hard)
Question Stats:
59% (02:53) correct
41%
(03:20)
wrong
based on 154
sessions
History
Date
Time
Result
Not Attempted Yet
The local orchestra is holding tryouts. Six violinists are competing for 2 positions, 5 pianists are competing for 1 position, and 8 trombonists are competing for 4 positions. If Suzette and Monica are 2 of the 8 trombonists and if the two women can never be in the orchestra together, how many different personnel configurations for the orchestras are possible?
A. 4,125
B. 4,150
C. 4,175
D. 4,200
E. 4,225
_________________
A. 4,125
B. 4,150
C. 4,175
D. 4,200
E. 4,225
_________________
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Total number of ways of selecting the local orchestra = 6C2 * 5C1 * 8C4
Number of ways of selecting the orchestra by having both Suzette and Monica = 6C2 * 5C1 * 6C2
Number of ways of selecting the orchestra such that Suzette and Monica are not together = 6C2 * 5C1 (8C4 - 6C2) = 75 * 55 = 4125
Answer: A
Number of ways of selecting the orchestra by having both Suzette and Monica = 6C2 * 5C1 * 6C2
Number of ways of selecting the orchestra such that Suzette and Monica are not together = 6C2 * 5C1 (8C4 - 6C2) = 75 * 55 = 4125
Answer: A
RatneshS
Joined: 14 May 2014
Last visit: 13 Jul 2017
Posts: 49
Location: India
Schools: Mannheim"17
GMAT 1: 680 Q49 V31
GPA: 3.44
Vyshak
Hi
Please tell the difference between yours and mine approach
6C2.5C1(6C4 + 2C1.7C3)
RatneshS
Hi,
In your approach you have selected either Suzette or Monica (2C1) which is fine. However you have chosen 7 members(7C3) when there are only 6 members remaining, resulting in a wrong answer. 2C1 * 6C3 should provide you the correct answer.
In your approach the correct answer should be read as: 6C2.5C1(6C4 + 2C1.6C3)
Posts: 297
Hi Bunuel Sir chetan2u
Below are two approaches but I am unable to recognize the mistake in the second.
Approach 1 : 6C2 (Violins) * 5C1 (Piano) * (8C4-6C2) for trombonists - 4125 (correct answer)
Approach 2 :
6C2 (Violins) * 5C1 (Piano)
Now, for trombonists, I have 3 cases.
Case 1 : M is there, S is not there - 7C3 = 35
Case 2 : S is there, M is not there in the team - 7C3 = 35
Case 3 : Both are not there in the team = 6C4= 15
This gives me 85
So, total cases = 6C2 (Violins) * 5C1 (Piano) * 85 = 6375 (Incorrect)
I have been trying and trying since yesterday but unable to understand what the flaw is with approach #2
Below are two approaches but I am unable to recognize the mistake in the second.
Approach 1 : 6C2 (Violins) * 5C1 (Piano) * (8C4-6C2) for trombonists - 4125 (correct answer)
Approach 2 :
6C2 (Violins) * 5C1 (Piano)
Now, for trombonists, I have 3 cases.
Case 1 : M is there, S is not there - 7C3 = 35
Case 2 : S is there, M is not there in the team - 7C3 = 35
Case 3 : Both are not there in the team = 6C4= 15
This gives me 85
So, total cases = 6C2 (Violins) * 5C1 (Piano) * 85 = 6375 (Incorrect)
I have been trying and trying since yesterday but unable to understand what the flaw is with approach #2
carouselambra
You are wrong in the coloured portion.
If M is already there, then you have to select 4-1 or 3. From how many?: M is selected and S cannot be there, so out of remaining 6.=> 6C3=20
Similarly for S there but M not there: 20 cases
None of the two there => 6C4=15
Total: 20+20+15=55
Answer will again be 4125
_________________
Posts: 297
chetan2u
Ah. This was just so obvious
Thank you Sir!
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