Venn Diagram at least one course











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At the music academy, there are 43 students taking piano, 57 students taking violin, 29 students taking guitar, and 18 taking flute. There are 10 students in any two of these courses, 5 students in any three of them and 2 taking all courses. How many students are taking at least one course at the music academy?



It seems a bit too easy to just add up the individual courses, 43+57+29+18. My logic is that the 10 students taking 2 or more courses, the 5 in any three, and the 2 taking all the courses don't matter, since they are included in at least one course already.



edit: The fact that some students are taking more than one of the courses would mean I would have to subtract those students from the total number of students in each course, making the answer is 43+57+29+18-10-5-2?



edit2: So rereading the question I came up with this.attempted venn diagram After subtracting the 40 students taking at least 2 courses, the 5 students taking at least 3, and the 2 students taking all, I've got 85 students taking at least 1 course.










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    But, among the 43 students taking piano, some of those are also taking violin. Approach via the Inclusion-Exclusion Principle.
    – JMoravitz
    Oct 28 '16 at 23:01















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At the music academy, there are 43 students taking piano, 57 students taking violin, 29 students taking guitar, and 18 taking flute. There are 10 students in any two of these courses, 5 students in any three of them and 2 taking all courses. How many students are taking at least one course at the music academy?



It seems a bit too easy to just add up the individual courses, 43+57+29+18. My logic is that the 10 students taking 2 or more courses, the 5 in any three, and the 2 taking all the courses don't matter, since they are included in at least one course already.



edit: The fact that some students are taking more than one of the courses would mean I would have to subtract those students from the total number of students in each course, making the answer is 43+57+29+18-10-5-2?



edit2: So rereading the question I came up with this.attempted venn diagram After subtracting the 40 students taking at least 2 courses, the 5 students taking at least 3, and the 2 students taking all, I've got 85 students taking at least 1 course.










share|cite|improve this question




















  • 1




    But, among the 43 students taking piano, some of those are also taking violin. Approach via the Inclusion-Exclusion Principle.
    – JMoravitz
    Oct 28 '16 at 23:01













up vote
0
down vote

favorite









up vote
0
down vote

favorite











At the music academy, there are 43 students taking piano, 57 students taking violin, 29 students taking guitar, and 18 taking flute. There are 10 students in any two of these courses, 5 students in any three of them and 2 taking all courses. How many students are taking at least one course at the music academy?



It seems a bit too easy to just add up the individual courses, 43+57+29+18. My logic is that the 10 students taking 2 or more courses, the 5 in any three, and the 2 taking all the courses don't matter, since they are included in at least one course already.



edit: The fact that some students are taking more than one of the courses would mean I would have to subtract those students from the total number of students in each course, making the answer is 43+57+29+18-10-5-2?



edit2: So rereading the question I came up with this.attempted venn diagram After subtracting the 40 students taking at least 2 courses, the 5 students taking at least 3, and the 2 students taking all, I've got 85 students taking at least 1 course.










share|cite|improve this question















At the music academy, there are 43 students taking piano, 57 students taking violin, 29 students taking guitar, and 18 taking flute. There are 10 students in any two of these courses, 5 students in any three of them and 2 taking all courses. How many students are taking at least one course at the music academy?



It seems a bit too easy to just add up the individual courses, 43+57+29+18. My logic is that the 10 students taking 2 or more courses, the 5 in any three, and the 2 taking all the courses don't matter, since they are included in at least one course already.



edit: The fact that some students are taking more than one of the courses would mean I would have to subtract those students from the total number of students in each course, making the answer is 43+57+29+18-10-5-2?



edit2: So rereading the question I came up with this.attempted venn diagram After subtracting the 40 students taking at least 2 courses, the 5 students taking at least 3, and the 2 students taking all, I've got 85 students taking at least 1 course.







probability combinatorics discrete-mathematics






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edited Oct 28 '16 at 23:21

























asked Oct 28 '16 at 22:55









dan

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  • 1




    But, among the 43 students taking piano, some of those are also taking violin. Approach via the Inclusion-Exclusion Principle.
    – JMoravitz
    Oct 28 '16 at 23:01














  • 1




    But, among the 43 students taking piano, some of those are also taking violin. Approach via the Inclusion-Exclusion Principle.
    – JMoravitz
    Oct 28 '16 at 23:01








1




1




But, among the 43 students taking piano, some of those are also taking violin. Approach via the Inclusion-Exclusion Principle.
– JMoravitz
Oct 28 '16 at 23:01




But, among the 43 students taking piano, some of those are also taking violin. Approach via the Inclusion-Exclusion Principle.
– JMoravitz
Oct 28 '16 at 23:01










2 Answers
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Let $P,V,G,F$ represent the sets of students taking piano (and possibly others), violin (and possibly others).



Inclusion-exclusion tells us that:



$|Pcup Vcup Gcup F| = |P|+|V|+|G|+|F|-|Pcap V|-|Pcap G|-|Pcap F|-|Vcap G|-|Vcap F| - |Gcap F|$



$~~~~~~~dots+|Pcap Vcap G|+|Pcap Vcap F|+|Pcap Gcap F|+|Vcap Gcap F| - |Pcap Vcap Gcap F|$






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    You might find this 4-set Venn layout useful (one of several options):



    enter image description here



    Attribution




    By RupertMillard - Own work by uploader - I made this in Inkscape from my own recollection of Venn's construction, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=6034493




    You'll find 4 distinct regions of exactly-3-set overlap and 6 regions of exactly-2-set overlap. As far as I can tell, the question dictates that the populations in each distinct overlap region look like this:



    enter image description here



    The non-overlap regions of each set can then be easily calculated, and the total student count otained.






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      2 Answers
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      2 Answers
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      active

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      Let $P,V,G,F$ represent the sets of students taking piano (and possibly others), violin (and possibly others).



      Inclusion-exclusion tells us that:



      $|Pcup Vcup Gcup F| = |P|+|V|+|G|+|F|-|Pcap V|-|Pcap G|-|Pcap F|-|Vcap G|-|Vcap F| - |Gcap F|$



      $~~~~~~~dots+|Pcap Vcap G|+|Pcap Vcap F|+|Pcap Gcap F|+|Vcap Gcap F| - |Pcap Vcap Gcap F|$






      share|cite|improve this answer

























        up vote
        0
        down vote













        Let $P,V,G,F$ represent the sets of students taking piano (and possibly others), violin (and possibly others).



        Inclusion-exclusion tells us that:



        $|Pcup Vcup Gcup F| = |P|+|V|+|G|+|F|-|Pcap V|-|Pcap G|-|Pcap F|-|Vcap G|-|Vcap F| - |Gcap F|$



        $~~~~~~~dots+|Pcap Vcap G|+|Pcap Vcap F|+|Pcap Gcap F|+|Vcap Gcap F| - |Pcap Vcap Gcap F|$






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          Let $P,V,G,F$ represent the sets of students taking piano (and possibly others), violin (and possibly others).



          Inclusion-exclusion tells us that:



          $|Pcup Vcup Gcup F| = |P|+|V|+|G|+|F|-|Pcap V|-|Pcap G|-|Pcap F|-|Vcap G|-|Vcap F| - |Gcap F|$



          $~~~~~~~dots+|Pcap Vcap G|+|Pcap Vcap F|+|Pcap Gcap F|+|Vcap Gcap F| - |Pcap Vcap Gcap F|$






          share|cite|improve this answer












          Let $P,V,G,F$ represent the sets of students taking piano (and possibly others), violin (and possibly others).



          Inclusion-exclusion tells us that:



          $|Pcup Vcup Gcup F| = |P|+|V|+|G|+|F|-|Pcap V|-|Pcap G|-|Pcap F|-|Vcap G|-|Vcap F| - |Gcap F|$



          $~~~~~~~dots+|Pcap Vcap G|+|Pcap Vcap F|+|Pcap Gcap F|+|Vcap Gcap F| - |Pcap Vcap Gcap F|$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 28 '16 at 23:06









          JMoravitz

          46.1k33784




          46.1k33784






















              up vote
              0
              down vote













              You might find this 4-set Venn layout useful (one of several options):



              enter image description here



              Attribution




              By RupertMillard - Own work by uploader - I made this in Inkscape from my own recollection of Venn's construction, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=6034493




              You'll find 4 distinct regions of exactly-3-set overlap and 6 regions of exactly-2-set overlap. As far as I can tell, the question dictates that the populations in each distinct overlap region look like this:



              enter image description here



              The non-overlap regions of each set can then be easily calculated, and the total student count otained.






              share|cite|improve this answer

























                up vote
                0
                down vote













                You might find this 4-set Venn layout useful (one of several options):



                enter image description here



                Attribution




                By RupertMillard - Own work by uploader - I made this in Inkscape from my own recollection of Venn's construction, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=6034493




                You'll find 4 distinct regions of exactly-3-set overlap and 6 regions of exactly-2-set overlap. As far as I can tell, the question dictates that the populations in each distinct overlap region look like this:



                enter image description here



                The non-overlap regions of each set can then be easily calculated, and the total student count otained.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  You might find this 4-set Venn layout useful (one of several options):



                  enter image description here



                  Attribution




                  By RupertMillard - Own work by uploader - I made this in Inkscape from my own recollection of Venn's construction, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=6034493




                  You'll find 4 distinct regions of exactly-3-set overlap and 6 regions of exactly-2-set overlap. As far as I can tell, the question dictates that the populations in each distinct overlap region look like this:



                  enter image description here



                  The non-overlap regions of each set can then be easily calculated, and the total student count otained.






                  share|cite|improve this answer












                  You might find this 4-set Venn layout useful (one of several options):



                  enter image description here



                  Attribution




                  By RupertMillard - Own work by uploader - I made this in Inkscape from my own recollection of Venn's construction, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=6034493




                  You'll find 4 distinct regions of exactly-3-set overlap and 6 regions of exactly-2-set overlap. As far as I can tell, the question dictates that the populations in each distinct overlap region look like this:



                  enter image description here



                  The non-overlap regions of each set can then be easily calculated, and the total student count otained.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 2 '16 at 18:27









                  Joffan

                  32.1k43169




                  32.1k43169






























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