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From ETS Major Field Test in Mathematics

A student is given an exam consisting of 8 essay questions divided into 4 groups of 2 questions each. The student is required to select a set of 6 questions to answer, including at least 1 question from each of the 4 groups. How many sets of questions satisfy this requirement?

I'm thinking $$\binom{2}{1}^4 \binom{4}{2}$$

because we have to pick 1 from each of the 4 groups of 2 and then from the remaining 4 questions we pick 2.

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    $\begingroup$ You're overcounting a little: Suppose the student picks 1a,2a,3a,4a and then additionally chooses 1b and 2b, this is counted differently to picking 1b,2b,3a,4a and then 1a and 2a in your method. You can either account for the overcounting or count differently... Hint: first pick which groups contain two answered questions. $\endgroup$ – B. Mehta Jul 30 '18 at 22:46
  • $\begingroup$ @B.Mehta Oh right. Thought of it but forgot to mention. Thanks! Your suggestion? $\endgroup$ – BCLC Jul 30 '18 at 22:51
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    $\begingroup$ Either option works; if anything you should try both for the practice, and use them to check against each other :) My first thought on reading the question was to count differently. $\endgroup$ – B. Mehta Jul 30 '18 at 22:52
  • $\begingroup$ @B.Mehta Which option? Try both of what? I posted an answer $\endgroup$ – BCLC Jul 30 '18 at 22:57
  • $\begingroup$ @B.Mehta Do you think there's a way to get from 96 to 24? Perhaps dividing by 4 because order doesn't matter or something? Then again I think that would be 4! not 4. Or perhaps dividing by 2!2! or something $\endgroup$ – BCLC Jul 30 '18 at 22:58
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The student can choose two questions to omit, not both i the same group. There are $8$ options for the first question, and then there are $6$ options left for the second question. Of course the order in which the questions are chosrn doesn't matter, so we get $$\frac{8\times6}{2}=24$$ options.

Alternatively, the student can choose two groups to omit a question from, and then one question from each group. This way we get $$\binom{4}{2}\binom{2}{1}\binom{2}{1}=6\times2\times2=24$$ options.

Finally, in line with your own approach, we can first choose one question from each group. We can indeed do so in $16$ ways. Then we can choose two more questions, indeed in $6$ ways. But now we have overcounted; we reach the same set of questions if we had first chosen other questions in the two groups we ended up choosing both questions from. In how many ways could we have chosen questions from these two groups? A total of $4$ ways. So by this method we have coumted each set of questions $4$ times. Hence the total number of options is $\frac{96}{4}=24$.

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  • $\begingroup$ Thanks Servaes! Your alternative seems to closely match my original approach as compared to your original or my alternative. Thus, could my original approach be modified somehow? Like could we somehow divide 4 or 2!2! to 96 to get 24? $\endgroup$ – BCLC Jul 30 '18 at 23:04
  • $\begingroup$ Thanks for addressing original! $\endgroup$ – BCLC Jul 31 '18 at 2:09
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I'm thinking $$\binom{2}{1}^4 \binom{4}{2}$$

because we have to pick 1 from each of the 4 groups of 2 and then from the remaining 4 questions we pick 2.

You are overcounting here: there will be two groups from which you end up picking both questions, and there is of course only one way to pick two questions from a group of two: pick both!

However, by first picking $1$ out of two (which can be done in two ways), and then picking the other one, your method ends up counting two ways to pick both questions from a group ... but those two ways are really just two different orderings of the group of two, and ordering is not a consideration for this question. Therefore, you overcount by a factor of $2$ for each of the groups where you end up picking both questions.

Finally, since there are two such groups, you overcount by a factor of $4$.

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  • $\begingroup$ Thanks for addressing original Bram28! $\endgroup$ – BCLC Jul 31 '18 at 2:10
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    $\begingroup$ @BCLC Sure thing! I always prefer to comment on the method used by the OP rather than proposing my own method. $\endgroup$ – Bram28 Jul 31 '18 at 10:54
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$$\binom{8}{6}=28$$

Then let's take away the choices where we don't cover all 4 groups. There are only 4. Why? To not cover all 4 groups but to choose 6 means to pick all questions but 2 questions of the same group. That is, from 4 groups, pick 1 group to exclude or equivalently pick 3 groups to include $$\binom{4}{3}=\binom{4}{1}=4$$

$$\therefore \binom{8}{6} - \binom{4}{3} = \binom{8}{6} - \binom{4}{1} = 28-4=24$$

Please suggest how my original approach could have been improved. I think the essence of the original approach is that we first pick 1 from each, and I guess that could be done in 16 ways. I'm just not sure how to reach 24 from there. We could still go to 96 but then I'm not sure how to reach 24 from 96. I guess something about how order doesn't matter twice so we would divide 96 by 2!2!.

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Because we only have four groups and you need to pick up 6 balls which means you need to pick up two groups of balls and one ball each from the other two groups so the final result would be

${4\choose2} \times 2 \times 2$ which is 24.

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  • $\begingroup$ Ah, pigeonholed. Thanks xiaonan! $\endgroup$ – BCLC Jul 31 '18 at 2:10
  • $\begingroup$ You meant to say "you need to pick up two groups of balls and one ball each from the other two groups ...". $\endgroup$ – N. F. Taussig Jul 31 '18 at 9:42
  • $\begingroup$ Thanks Taussig! $\endgroup$ – Xiaonan Jul 31 '18 at 18:35

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