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.
I solved it using the following method
3C1 * 3C1 * 6C1 * 6C1 * 14C1
1st q 2nd q 3rd q 4th q 5th q
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!!!