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This question has disturbed several nights of my sleep.

There are 3 sections in a question paper and each section contains 5 questions. A candidate has to answer a total of 5 questions, choosing at least one question from each section. Then the number of ways, in which the candidate can choose the questions, is :


My Approach:

Since it is compulsory to select at least one question from 3 section, I selected it by doing $\binom{5}{1}\cdot\binom{5}{1}\cdot\binom{5}{1}$

Now there are 12 questions left in total. Since there is no restrictions in selecting the remaining questions, I did it by thinking it to be a pool of 12 questions from where I picked up 2 questions to fulfill my requirement. I did it by doing $\binom{12}{2}$.

So the total number of ways of selecting 5 questions are $\binom{5}{1}\cdot\binom{5}{1}\cdot\binom{5}{1}\cdot\binom{12}{2} = 8250$

But the correct answer is $\boxed{2250}$

I absolutely cannot figure out where have I gone wrong or how did I overcounted!

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3 Answers 3

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Your way of counting is correct if you are required to mark one of your selected questions from each section with a star. But this overcounts because we don't care which question is your "starred" representative of the section.

For simplicity, suppose the 15 questions are numbered 1 to 15, with the first 5 questions being in the first section, and so on.

One outcome is to choose questions 1, 2, 3, 6, 11. However, you count this three times:

  • 1*, 2, 3, 6*, 11*
  • 1, 2*, 3, 6*, 11*
  • 1, 2, 3*, 6*, 11*

Another outcome is to choose questions 1, 2, 6, 7, 11. However, you count this four times:

  • 1*, 2, 6*, 7, 11*
  • 1*, 2, 6, 7*, 11*
  • 1, 2*, 6*, 7, 11*
  • 1, 2*, 6, 7*, 11*

In general for each actual outcome, you overcount by either 3 times or 4 times, which explains why $8250/2250 = 3.\bar{6}$ is between 3 and 4.

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  • $\begingroup$ I can sleep well now $\endgroup$
    – ADITYA DAS
    Commented Jun 25 at 18:12
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@angryavian explained it well, but I am putting similar question link here to expand our understanding.

When we come to how to solve this question, the answer is Inclusion-exclusion principle such that (all cases without restriction) - ( one section is not included) + ( two section are not included) - ( three section are not inluded)

$$\binom{15}{5}-3\binom{10}{5}+3\binom{5}{5}-1\binom{0}{5}=3003-3(252)+3(1)-0=2250$$

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Overcounting by your method has been nicely explained by angryavian

To get the correct count, you could use the format

[Select sectionwise numbers] x [Permute sections]

$= \left[\binom52\binom52\binom51 *\frac{3!}{2!}\right] + \left[\binom53\binom51\binom51 *\frac{3!}{2!}\right] = [1500+750] =2250$

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