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# The local orchestra is holding tryouts. Six violinists are competing

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The local orchestra is holding tryouts. Six violinists are competing [#permalink]
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RatneshS wrote:
Vyshak wrote:
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

Hi

Please tell the difference between yours and mine approach

6C2.5C1(6C4 + 2C1.7C3)

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.

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The local orchestra is holding tryouts. Six violinists are competing [#permalink]
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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Re: The local orchestra is holding tryouts. Six violinists are competing [#permalink]
carouselambra wrote:
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
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Re: The local orchestra is holding tryouts. Six violinists are competing [#permalink]
chetan2u wrote:
carouselambra wrote:
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

Ah. This was just so obvious
Thank you Sir!
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Re: The local orchestra is holding tryouts. Six violinists are competing [#permalink]
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Re: The local orchestra is holding tryouts. Six violinists are competing [#permalink]
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