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On a class test there are 5 questions. One question has been [#permalink]

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27 Mar 2013, 11:41

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On a class test there are 5 questions. One question has been taken from each of four chapters. The first two chapters have 3 questions each, the last two chapters have 6 questions each. The fourth question can be picked from any of the four chapters. How many different question papers could have been prepared?

Re: On a class test there are 5 questions. One question has been [#permalink]

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27 Mar 2013, 13:50

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Are you sure of your answers?

I tried to figure it out but 4536 is my result, if some expert could help here... My way: number of ways to pick one from chapter I and II = 3C1 = 3 number of ways to pick one from chapter III and IV = 6C1 = 6 number of ways to pick one from the remaining 14 = 14C1 = 14 So my solution is 3*3*6*6*14=4536
_________________

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Re: On a class test there are 5 questions. One question has been [#permalink]

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29 Mar 2013, 14:24

Zarrolou wrote:

Are you sure of your answers?

I tried to figure it out but 4536 is my result, if some expert could help here... My way: number of ways to pick one from chapter I and II = 3C1 = 3 number of ways to pick one from chapter III and IV = 6C1 = 6 number of ways to pick one from the remaining 14 = 14C1 = 14 So my solution is 3*3*6*6*14=4536

+1 for 4536. please check and confirm the answer choices. thanks

Re: On a class test there are 5 questions. One question has been [#permalink]

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15 Sep 2013, 01:07

I feel like 3.3.6.6.14 is not the right answer.

Imagine the first textbook has questions Q1, Q2, Q3. Now we can choose one of these, so it's a multiplier of 3. Let's say we picked Q1. Once we've finished picking the 4 questions and now it's time to pick the final question, imagine we pick from the first textbook. Since we already picked Q1, now we can only pick Q2 and Q3. This is a multiplier of 2. Let's imagine we picked Q2.

However, imagine another situation where we picked Q2 first, and then in the fifth round we can pick Q1 or Q3. It's still a multiplier of 2. Let's imagine we picked Q1.

However, picking Q1 first then Q2 later, or Q2 first and Q1 later, are the same combinations. But by using the 14- multipliers, we are not accounting for this.

Imagine the first textbook has questions Q1, Q2, Q3. Now we can choose one of these, so it's a multiplier of 3. Let's say we picked Q1. Once we've finished picking the 4 questions and now it's time to pick the final question, imagine we pick from the first textbook. Since we already picked Q1, now we can only pick Q2 and Q3. This is a multiplier of 2. Let's imagine we picked Q2.

However, imagine another situation where we picked Q2 first, and then in the fifth round we can pick Q1 or Q3. It's still a multiplier of 2. Let's imagine we picked Q1.

However, picking Q1 first then Q2 later, or Q2 first and Q1 later, are the same combinations. But by using the 14- multipliers, we are not accounting for this.

I think the OA is wrong as well...

That's correct.

On a class test there are 5 questions. One question has been taken from each of four chapters. The first two chapters have 3 questions each, the last two chapters have 6 questions each. The fourth question can be picked from any of the four chapters. How many different question papers could have been prepared? A. 540 B. 1260 C. 1080 D. 400 E. 4860

We can pick 2 questions from either of the four chapters.

If 2 questions are picked from the first chapter: \(C^2_3*3*6*6\);

If 2 questions are picked from the second chapter: \(3*C^2_3*6*6\);

If 2 questions are picked from the third chapter: \(3*3*C^2_6*6\);

If 2 questions are picked from the third chapter: \(3*3*6*C^2_6\).

Total = 2,268.

There is no correct answer among the options.
_________________

Re: On a class test there are 5 questions. One question has been [#permalink]

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15 Sep 2013, 03:07

Bunuel wrote:

jubinell wrote:

I feel like 3.3.6.6.14 is not the right answer.

Imagine the first textbook has questions Q1, Q2, Q3. Now we can choose one of these, so it's a multiplier of 3. Let's say we picked Q1. Once we've finished picking the 4 questions and now it's time to pick the final question, imagine we pick from the first textbook. Since we already picked Q1, now we can only pick Q2 and Q3. This is a multiplier of 2. Let's imagine we picked Q2.

However, imagine another situation where we picked Q2 first, and then in the fifth round we can pick Q1 or Q3. It's still a multiplier of 2. Let's imagine we picked Q1.

However, picking Q1 first then Q2 later, or Q2 first and Q1 later, are the same combinations. But by using the 14- multipliers, we are not accounting for this.

I think the OA is wrong as well...

That's correct.

On a class test there are 5 questions. One question has been taken from each of four chapters. The first two chapters have 3 questions each, the last two chapters have 6 questions each. The fourth question can be picked from any of the four chapters. How many different question papers could have been prepared? A. 540 B. 1260 C. 1080 D. 400 E. 4860

We can pick 2 questions from either of the four chapters.

If 2 questions are picked from the first chapter: \(C^2_3*3*6*6\);

If 2 questions are picked from the second chapter: \(3*C^2_3*6*6\);

If 2 questions are picked from the third chapter: \(3*3*C^2_6*6\);

If 2 questions are picked from the third chapter: \(3*3*6*C^2_6\).

Total = 2,268.

There is no correct answer among the options.

Hi Bunuel,

I understood the way you have solved, but could you please help me understanding where I am going wrong.

Imagine the first textbook has questions Q1, Q2, Q3. Now we can choose one of these, so it's a multiplier of 3. Let's say we picked Q1. Once we've finished picking the 4 questions and now it's time to pick the final question, imagine we pick from the first textbook. Since we already picked Q1, now we can only pick Q2 and Q3. This is a multiplier of 2. Let's imagine we picked Q2.

However, imagine another situation where we picked Q2 first, and then in the fifth round we can pick Q1 or Q3. It's still a multiplier of 2. Let's imagine we picked Q1.

However, picking Q1 first then Q2 later, or Q2 first and Q1 later, are the same combinations. But by using the 14- multipliers, we are not accounting for this.

I think the OA is wrong as well...

That's correct.

On a class test there are 5 questions. One question has been taken from each of four chapters. The first two chapters have 3 questions each, the last two chapters have 6 questions each. The fourth question can be picked from any of the four chapters. How many different question papers could have been prepared? A. 540 B. 1260 C. 1080 D. 400 E. 4860

We can pick 2 questions from either of the four chapters.

If 2 questions are picked from the first chapter: \(C^2_3*3*6*6\);

If 2 questions are picked from the second chapter: \(3*C^2_3*6*6\);

If 2 questions are picked from the third chapter: \(3*3*C^2_6*6\);

If 2 questions are picked from the third chapter: \(3*3*6*C^2_6\).

Total = 2,268.

There is no correct answer among the options.

Hi Bunuel,

I understood the way you have solved, but could you please help me understanding where I am going wrong.

Or consider this. With your solution we can get: A1, B1, C1, D1, and then A2. But we can also get: A2, B1, C1, D1, and then A1, which is basically the same set.
_________________

Re: On a class test there are 5 questions. One question has been [#permalink]

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15 Sep 2013, 11:00

shameekv wrote:

Bunuel wrote:

jubinell wrote:

I feel like 3.3.6.6.14 is not the right answer.

Imagine the first textbook has questions Q1, Q2, Q3. Now we can choose one of these, so it's a multiplier of 3. Let's say we picked Q1. Once we've finished picking the 4 questions and now it's time to pick the final question, imagine we pick from the first textbook. Since we already picked Q1, now we can only pick Q2 and Q3. This is a multiplier of 2. Let's imagine we picked Q2.

However, imagine another situation where we picked Q2 first, and then in the fifth round we can pick Q1 or Q3. It's still a multiplier of 2. Let's imagine we picked Q1.

However, picking Q1 first then Q2 later, or Q2 first and Q1 later, are the same combinations. But by using the 14- multipliers, we are not accounting for this.

I think the OA is wrong as well...

That's correct.

On a class test there are 5 questions. One question has been taken from each of four chapters. The first two chapters have 3 questions each, the last two chapters have 6 questions each. The fourth question can be picked from any of the four chapters. How many different question papers could have been prepared? A. 540 B. 1260 C. 1080 D. 400 E. 4860

We can pick 2 questions from either of the four chapters.

If 2 questions are picked from the first chapter: \(C^2_3*3*6*6\);

If 2 questions are picked from the second chapter: \(3*C^2_3*6*6\);

If 2 questions are picked from the third chapter: \(3*3*C^2_6*6\);

If 2 questions are picked from the third chapter: \(3*3*6*C^2_6\).

Total = 2,268.

There is no correct answer among the options.

Hi Bunuel,

I understood the way you have solved, but could you please help me understanding where I am going wrong.

14C1 is because we are left with 2+2+5+5 questions in each of the chapters respectively and that we have to chose the last one from those remaining.

When I solve this I get double of 2268 i.e. 4536...

Thanks in advance!!!

Hi, Well I am still in confusion..

What if the question had asked if there were 6 questions to be picked and last 2 can be picked from any of those questions remaining? And also if the last 2 can be picked only from one of the chapters.

Are we trying to divide the term 4536 by 2! since 2 questions represent same set from 1 chapter?

Could you please elaborate on this? I am trying to understand the concept here..

On a class test there are 5 questions. One question has been taken from each of four chapters. The first two chapters have 3 questions each, the last two chapters have 6 questions each. The fourth question can be picked from any of the four chapters. How many different question papers could have been prepared? A. 540 B. 1260 C. 1080 D. 400 E. 4860

We can pick 2 questions from either of the four chapters.

If 2 questions are picked from the first chapter: \(C^2_3*3*6*6\);

If 2 questions are picked from the second chapter: \(3*C^2_3*6*6\);

If 2 questions are picked from the third chapter: \(3*3*C^2_6*6\);

If 2 questions are picked from the third chapter: \(3*3*6*C^2_6\).

Total = 2,268.

There is no correct answer among the options.

Hi Bunuel,

I understood the way you have solved, but could you please help me understanding where I am going wrong.

14C1 is because we are left with 2+2+5+5 questions in each of the chapters respectively and that we have to chose the last one from those remaining.

When I solve this I get double of 2268 i.e. 4536...

Thanks in advance!!!

Hi, Well I am still in confusion..

What if the question had asked if there were 6 questions to be picked and last 2 can be picked from any of those questions remaining? And also if the last 2 can be picked only from one of the chapters.

Are we trying to divide the term 4536 by 2! since 2 questions represent same set from 1 chapter?

Could you please elaborate on this? I am trying to understand the concept here..

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