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The local orchestra is holding tryouts. Six violinists are competing 08 Apr 2016, 02:56
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59% (02:53) correct 41% (03:20) wrong based on 153 sessions

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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
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A
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The local orchestra is holding tryouts. Six violinists are competing 08 Apr 2016, 21:40
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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
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The local orchestra is holding tryouts. Six violinists are competing 09 Apr 2016, 07:09
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Vyshak
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
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Hi

Please tell the difference between yours and mine approach

6C2.5C1(6C4 + 2C1.7C3)
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The local orchestra is holding tryouts. Six violinists are competing 09 Apr 2016, 07:22
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Vyshak
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


Hi

Please tell the difference between yours and mine approach

6C2.5C1(6C4 + 2C1.7C3)
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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)
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The local orchestra is holding tryouts. Six violinists are competing 11 Aug 2021, 20:24
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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
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The local orchestra is holding tryouts. Six violinists are competing 11 Aug 2021, 20:56
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carouselambra
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
Show more

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
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The local orchestra is holding tryouts. Six violinists are competing 11 Aug 2021, 21:01
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chetan2u
carouselambra
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

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
Show more

Ah. This was just so obvious :(
Thank you Sir!
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The local orchestra is holding tryouts. Six violinists are competing 10 Jul 2023, 07:05
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