Common doubt: Why my process is wrong?











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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. I 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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    You have made the last paragraph into "spoiler text" that is blanked out until the user navigates their pointer over it. Did you really want that effect?
    – David K
    1 hour ago















up vote
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down vote

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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. I 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.











share|cite|improve this question




















  • 1




    You have made the last paragraph into "spoiler text" that is blanked out until the user navigates their pointer over it. Did you really want that effect?
    – David K
    1 hour ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











Question



A question paper on mathematics contains $12$ questions divided into three parts A, B and C, each containing $4$ questions. I 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.











share|cite|improve this question















Question



A question paper on mathematics contains $12$ questions divided into three parts A, B and C, each containing $4$ questions. I 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.








combinatorics combinations






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edited 24 mins ago









N. F. Taussig

43.2k93254




43.2k93254










asked 1 hour ago









jayant98

36814




36814








  • 1




    You have made the last paragraph into "spoiler text" that is blanked out until the user navigates their pointer over it. Did you really want that effect?
    – David K
    1 hour ago














  • 1




    You have made the last paragraph into "spoiler text" that is blanked out until the user navigates their pointer over it. Did you really want that effect?
    – David K
    1 hour ago








1




1




You have made the last paragraph into "spoiler text" that is blanked out until the user navigates their pointer over it. Did you really want that effect?
– David K
1 hour ago




You have made the last paragraph into "spoiler text" that is blanked out until the user navigates their pointer over it. Did you really want that effect?
– David K
1 hour ago










3 Answers
3






active

oldest

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up vote
1
down vote



accepted










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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    up vote
    3
    down vote













    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.






    share|cite|improve this answer





















    • 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?
      – jayant98
      55 mins ago




















    up vote
    3
    down vote













    To select just three questions, one from each section, you are correct that there are $4times 4times 4$ 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
    $64times 36 = 2304$ 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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    • Oh. Understood your point. Thanks for clearing my doubt. Thank you very much!
      – jayant98
      52 mins ago










    • @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).
      – adhg
      36 mins ago











    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










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






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










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






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






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






        share|cite|improve this answer












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







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 36 mins ago









        N. F. Taussig

        43.2k93254




        43.2k93254






















            up vote
            3
            down vote













            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.






            share|cite|improve this answer





















            • 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?
              – jayant98
              55 mins ago

















            up vote
            3
            down vote













            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.






            share|cite|improve this answer





















            • 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?
              – jayant98
              55 mins ago















            up vote
            3
            down vote










            up vote
            3
            down vote









            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.






            share|cite|improve this answer












            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Bram28

            59.2k44186




            59.2k44186












            • 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?
              – jayant98
              55 mins ago




















            • 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?
              – jayant98
              55 mins ago


















            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?
            – jayant98
            55 mins ago






            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?
            – jayant98
            55 mins ago












            up vote
            3
            down vote













            To select just three questions, one from each section, you are correct that there are $4times 4times 4$ 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
            $64times 36 = 2304$ 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.






            share|cite|improve this answer





















            • Oh. Understood your point. Thanks for clearing my doubt. Thank you very much!
              – jayant98
              52 mins ago










            • @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).
              – adhg
              36 mins ago















            up vote
            3
            down vote













            To select just three questions, one from each section, you are correct that there are $4times 4times 4$ 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
            $64times 36 = 2304$ 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.






            share|cite|improve this answer





















            • Oh. Understood your point. Thanks for clearing my doubt. Thank you very much!
              – jayant98
              52 mins ago










            • @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).
              – adhg
              36 mins ago













            up vote
            3
            down vote










            up vote
            3
            down vote









            To select just three questions, one from each section, you are correct that there are $4times 4times 4$ 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
            $64times 36 = 2304$ 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.






            share|cite|improve this answer












            To select just three questions, one from each section, you are correct that there are $4times 4times 4$ 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
            $64times 36 = 2304$ 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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            David K

            52.1k340115




            52.1k340115












            • Oh. Understood your point. Thanks for clearing my doubt. Thank you very much!
              – jayant98
              52 mins ago










            • @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).
              – adhg
              36 mins ago


















            • Oh. Understood your point. Thanks for clearing my doubt. Thank you very much!
              – jayant98
              52 mins ago










            • @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).
              – adhg
              36 mins ago
















            Oh. Understood your point. Thanks for clearing my doubt. Thank you very much!
            – jayant98
            52 mins ago




            Oh. Understood your point. Thanks for clearing my doubt. Thank you very much!
            – jayant98
            52 mins ago












            @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).
            – adhg
            36 mins ago




            @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).
            – adhg
            36 mins ago


















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