Mixing
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.
probability combinatorics discrete-mathematics
asked Oct 28 '16 at 22:55
dandan
12
$endgroup$
add a comment |
$begingroup$
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
asked Oct 28 '16 at 22:55
dandan
12
$endgroup$
1
$begingroup$
But, among the 43 students taking piano, some of those are also taking violin. Approach via the Inclusion-Exclusion Principle.
$endgroup$
– JMoravitz
Oct 28 '16 at 23:01
add a comment |
$begingroup$
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
asked Oct 28 '16 at 22:55
dandan
12
$endgroup$
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
probability combinatorics discrete-mathematics
asked Oct 28 '16 at 22:55
dandan
12
asked Oct 28 '16 at 22:55
dandan
12
edited Oct 28 '16 at 23:21
dan
asked Oct 28 '16 at 22:55
dandan
12
asked Oct 28 '16 at 22:55
dandan
12
asked Oct 28 '16 at 22:55
dandan
12
12
1
$begingroup$
But, among the 43 students taking piano, some of those are also taking violin. Approach via the Inclusion-Exclusion Principle.
$endgroup$
– JMoravitz
Oct 28 '16 at 23:01
add a comment |
1
$begingroup$
But, among the 43 students taking piano, some of those are also taking violin. Approach via the Inclusion-Exclusion Principle.
$endgroup$
– JMoravitz
Oct 28 '16 at 23:01
1
1
$begingroup$
But, among the 43 students taking piano, some of those are also taking violin. Approach via the Inclusion-Exclusion Principle.
$endgroup$
– JMoravitz
Oct 28 '16 at 23:01
$begingroup$
But, among the 43 students taking piano, some of those are also taking violin. Approach via the Inclusion-Exclusion Principle.
$endgroup$
– JMoravitz
Oct 28 '16 at 23:01
add a comment |
2 Answers
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$begingroup$
Let $P,V,G,F$ represent the sets of students taking piano (and possibly others), violin (and possibly others).
Inclusion-exclusion tells us that:
$|Pcup Vcup Gcup F| = |P|+|V|+|G|+|F|-|Pcap V|-|Pcap G|-|Pcap F|-|Vcap G|-|Vcap F| - |Gcap F|$
$~~~~~~~dots+|Pcap Vcap G|+|Pcap Vcap F|+|Pcap Gcap F|+|Vcap Gcap F| - |Pcap Vcap Gcap F|$
answered Oct 28 '16 at 23:06
JMoravitzJMoravitz
46.6k33886
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add a comment |
$begingroup$
You might find this 4-set Venn layout useful (one of several options):
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:
The non-overlap regions of each set can then be easily calculated, and the total student count otained.
answered Nov 2 '16 at 18:27
JoffanJoffan
32.2k43269
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add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
Let $P,V,G,F$ represent the sets of students taking piano (and possibly others), violin (and possibly others).
Inclusion-exclusion tells us that:
$|Pcup Vcup Gcup F| = |P|+|V|+|G|+|F|-|Pcap V|-|Pcap G|-|Pcap F|-|Vcap G|-|Vcap F| - |Gcap F|$
$~~~~~~~dots+|Pcap Vcap G|+|Pcap Vcap F|+|Pcap Gcap F|+|Vcap Gcap F| - |Pcap Vcap Gcap F|$
answered Oct 28 '16 at 23:06
JMoravitzJMoravitz
46.6k33886
$endgroup$
add a comment |
$begingroup$
Let $P,V,G,F$ represent the sets of students taking piano (and possibly others), violin (and possibly others).
Inclusion-exclusion tells us that:
$|Pcup Vcup Gcup F| = |P|+|V|+|G|+|F|-|Pcap V|-|Pcap G|-|Pcap F|-|Vcap G|-|Vcap F| - |Gcap F|$
$~~~~~~~dots+|Pcap Vcap G|+|Pcap Vcap F|+|Pcap Gcap F|+|Vcap Gcap F| - |Pcap Vcap Gcap F|$
answered Oct 28 '16 at 23:06
JMoravitzJMoravitz
46.6k33886
$endgroup$
add a comment |
$begingroup$
Let $P,V,G,F$ represent the sets of students taking piano (and possibly others), violin (and possibly others).
Inclusion-exclusion tells us that:
$|Pcup Vcup Gcup F| = |P|+|V|+|G|+|F|-|Pcap V|-|Pcap G|-|Pcap F|-|Vcap G|-|Vcap F| - |Gcap F|$
$~~~~~~~dots+|Pcap Vcap G|+|Pcap Vcap F|+|Pcap Gcap F|+|Vcap Gcap F| - |Pcap Vcap Gcap F|$
answered Oct 28 '16 at 23:06
JMoravitzJMoravitz
46.6k33886
$endgroup$
Let $P,V,G,F$ represent the sets of students taking piano (and possibly others), violin (and possibly others).
Inclusion-exclusion tells us that:
$|Pcup Vcup Gcup F| = |P|+|V|+|G|+|F|-|Pcap V|-|Pcap G|-|Pcap F|-|Vcap G|-|Vcap F| - |Gcap F|$
$~~~~~~~dots+|Pcap Vcap G|+|Pcap Vcap F|+|Pcap Gcap F|+|Vcap Gcap F| - |Pcap Vcap Gcap F|$
answered Oct 28 '16 at 23:06
JMoravitzJMoravitz
46.6k33886
answered Oct 28 '16 at 23:06
JMoravitzJMoravitz
46.6k33886
answered Oct 28 '16 at 23:06
JMoravitzJMoravitz
46.6k33886
answered Oct 28 '16 at 23:06
JMoravitzJMoravitz
46.6k33886
46.6k33886
add a comment |
add a comment |
$begingroup$
You might find this 4-set Venn layout useful (one of several options):
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:
The non-overlap regions of each set can then be easily calculated, and the total student count otained.
answered Nov 2 '16 at 18:27
JoffanJoffan
32.2k43269
$endgroup$
add a comment |
$begingroup$
You might find this 4-set Venn layout useful (one of several options):
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:
The non-overlap regions of each set can then be easily calculated, and the total student count otained.
answered Nov 2 '16 at 18:27
JoffanJoffan
32.2k43269
$endgroup$
add a comment |
$begingroup$
You might find this 4-set Venn layout useful (one of several options):
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:
The non-overlap regions of each set can then be easily calculated, and the total student count otained.
answered Nov 2 '16 at 18:27
JoffanJoffan
32.2k43269
$endgroup$
You might find this 4-set Venn layout useful (one of several options):
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:
The non-overlap regions of each set can then be easily calculated, and the total student count otained.
answered Nov 2 '16 at 18:27
JoffanJoffan
32.2k43269
answered Nov 2 '16 at 18:27
JoffanJoffan
32.2k43269
answered Nov 2 '16 at 18:27
JoffanJoffan
32.2k43269
answered Nov 2 '16 at 18:27
JoffanJoffan
32.2k43269
32.2k43269
add a comment |
add a comment |
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1
$begingroup$
But, among the 43 students taking piano, some of those are also taking violin. Approach via the Inclusion-Exclusion Principle.
$endgroup$
– JMoravitz
Oct 28 '16 at 23:01