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Question

A question paper on mathematics contains $12$ questions divided into three parts A, B and C, each containing $4$ questions. In how many ways can an examinee answer $5$ questions, selecting at least one from each part.

Attempt

Firstly, I selected three questions (one from each part) and it can be done $4 \cdot 4 \cdot 4$ ways. And hence the remaining two positions for two questions can be given in $^9{\mathrm C}_2$ since there is no restrictions now. So, total ways is $36 \times 64=2304$.

But, in the answer given in the solution manual is $624$. And the process described is:

A    B    C

1    1    3

1    2    2

which is then arranged for part to be $3 \times (4 \times 4 \times 4 + 4 \times 6 \times 6) = 624$.

Why my process is incorrect? I understand the second solution but, unable to find any fault in my attempt. Please explain.

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    $\begingroup$ "The process described is:" Are you sure that table is the only description? I would have no idea what it means if I didn't already know how to find the answer. $\endgroup$ – JiK Dec 16 '18 at 23:33
  • $\begingroup$ @JiK I have copied as it was written in the solution manual. If you want to see the pic, I can upload it. $\endgroup$ – jayant98 Dec 17 '18 at 4:42
  • $\begingroup$ Can you give a more appropriate title maybe? $\endgroup$ – Asaf Karagila Dec 17 '18 at 15:58
  • $\begingroup$ @Asaf Well, I will think the title is making others tempting to see what's in it as it maintains somewhat secrecy and that increase the chance of getting this question answered and hence, it also increases the chance of getting more quality answer. And I would like this to remain as it is. $\endgroup$ – jayant98 Dec 17 '18 at 16:44
  • $\begingroup$ This is not a clickbait website. Please make your title better. $\endgroup$ – Asaf Karagila Dec 17 '18 at 16:48
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Let's compare your method with the correct solution.

Solution: As you noted, either three questions are selected from one section with one each from the other two sections or two questions each are selected from two sections and one is selected from the remaining section.

Three questions from one section and one question each from the other two sections: Select the section from which three questions are drawn, select three of the four questions from that section, and select one of the four questions from each of the other two sections. This can be done in $$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1}$$ ways.

Two questions each from two sections and one question from the remaining section: Select the two sections from which two questions are drawn, select two of the four questions from each of those sections, and select one of the four questions from the other section. This can be done in $$\binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}$$ ways.

Total: Since the two cases are mutually exclusive and exhaustive, there are $$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1} + \binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1} = 624$$ ways to select five questions so that at least one is drawn from each of the three sections.

Why your method is wrong?

You are counting each selection in which three questions are drawn from one section and one question is drawn from each of the other sections three times, once for each way you could designate one of the three questions as the question that is drawn from that section. For example, suppose questions $A_1, A_2, A_3, B_1, C_1$ are selected. You count this selection three times.

\begin{array}{c c} \text{designated questions} & \text{additional questions}\\ \hline A_1, B_1, C_1 & A_2, A_3\\ A_2, B_1, C_1 & A_1, A_3\\ A_3, B_1, C_1 & A_2, A_3 \end{array}

You are counting each selection in which two questions each are drawn from two sections and one question is drawn from the other section four times, two times for each way you could designate one of the two question from each section from which two questions are drawn as the question that is drawn from that section. For instance, if questions $A_1, A_2, B_1, B_2, C_1$ are drawn, your method counts this selection four times.

\begin{array}{c c} \text{designated questions} & \text{additional questions}\\ \hline A_1, B_1, C_1 & A_2, B_2\\ A_1, B_2, C_1 & A_2, B_1\\ A_2, B_1, C_1 & A_1, B_2\\ A_2, B_2, C_1 & A_1, B_1 \end{array}

Notice that $$\binom{3}{1}\color{red}{\binom{3}{1}}\binom{4}{3}\binom{4}{1}\binom{4}{1} + \binom{3}{2}\color{red}{\binom{2}{1}}\binom{4}{2}\color{red}{\binom{2}{1}}\binom{4}{2}\binom{4}{1} = 2304$$

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To select just three questions, one from each section, you are correct that there are $4\times 4\times 4$ ways to do it.

Now if you choose two more questions from the remaining $9,$ there are (using various notations) $9C2 = \binom92 = {}^9C_2 = 36$ possible ways to do that step. If you consider each resulting list of questions to be "different", then you would have $64\times 36 = 2304$ possible ways.

The thing is, as far as this question is concerned, not every sequence of questions you chose according to your method is "different" from every other. For example, choosing questions A1, B2, and C1 for the three questions (one from each section), and then A2 and A3 for the remaining two out of nine, gives the same result as choosing questions A3, B2, and C1 and then choosing A1 and A2 for the remaining two out of nine. In both cases the examinee answers A1, A2, A3, B2, and C1.

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  • $\begingroup$ Oh. Understood your point. Thanks for clearing my doubt. Thank you very much! $\endgroup$ – jayant98 Dec 16 '18 at 21:42
  • $\begingroup$ @David K you're right with the analysis with respect to the OP strategy. Question: how would you go around it? Can you show how would you solve it (the OP's answer = 624 make sense but I didn't get the strategy proposed on intuitive level). $\endgroup$ – adhg Dec 16 '18 at 21:58
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    $\begingroup$ @adhg N.F. Taussig's answer explains the reasoning behind the $624$ answer in detail. The key insight to me is that there are two fundamentally different ways to distribute the five responses: 3 in one section, 1 in each other section; or 2 in one section, 2 in another, and 1 in the remaining section. You work out the number of possibilities in each case; then, since there are no possibilities that could have gotten counted in both cases, you can just add the two cases together. $\endgroup$ – David K Dec 16 '18 at 23:15
  • $\begingroup$ @David K got it. $\endgroup$ – adhg Dec 16 '18 at 23:42
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The problem with your method is that you are overcounting!

For example: if you choose question 1 from part A as one of the four from part A, but then later choose question 2 from part A as one of the two leftover, then you could end up with the same set of questions had you first chosen question 2 from A as one of the four from A, and then later question 1 from A as one of the two leftover questions. But your method counts this as two separate sets.

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  • $\begingroup$ So you are saying that if I select A1 question first and then A2 as one of the last two questions then we have count as A1.... A2.. And A2..... A1.. Separately but, this is the same situation situation. Is this you wanted to say? $\endgroup$ – jayant98 Dec 16 '18 at 21:39
  • $\begingroup$ @jayant98 exactly! You got it. $\endgroup$ – Bram28 Dec 16 '18 at 22:52

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