Mixing
Number of students and how many are taking X class (Discrete Structures)
$begingroup$
So I have my first quiz tomorrow and want to get off on the good foot, but I'm studying some problems and this one is particularly confusing...
There are a group of 191 students, of which 10 are taking French,
business and music; 36 are taking French and business; 20 are taking
French and music; 18 are taking business and music; 65 are taking
French; 76 are taking business; and 63 are taking music.
1a. How many are taking French or business (or both)? 1b. How many are
taking music or French (or both) but not business?
It might just be all the numbers confusing me, but I'm kind of lost. For 1a: the answer I got was 177 but I don't think that can be right, as there are 63 taking music, which would push the total over 191...
I haven't even gotten through 1b, and I don't want to look at the answer just yet, as I'm trying to work my way through it. Could anyone provide a bit of help?
discrete-mathematics
asked Sep 3 '14 at 1:02
secondublysecondubly
2921418
$endgroup$
add a comment |
$begingroup$
So I have my first quiz tomorrow and want to get off on the good foot, but I'm studying some problems and this one is particularly confusing...
There are a group of 191 students, of which 10 are taking French,
business and music; 36 are taking French and business; 20 are taking
French and music; 18 are taking business and music; 65 are taking
French; 76 are taking business; and 63 are taking music.
1a. How many are taking French or business (or both)? 1b. How many are
taking music or French (or both) but not business?
It might just be all the numbers confusing me, but I'm kind of lost. For 1a: the answer I got was 177 but I don't think that can be right, as there are 63 taking music, which would push the total over 191...
I haven't even gotten through 1b, and I don't want to look at the answer just yet, as I'm trying to work my way through it. Could anyone provide a bit of help?
discrete-mathematics
asked Sep 3 '14 at 1:02
secondublysecondubly
2921418
$endgroup$
$begingroup$
Draw a venn diagram
$endgroup$
– voldemort
Sep 3 '14 at 1:03
$begingroup$
I don't think you were correct, please look at my answer for clarification. The number of music students doesn't affect the value in 1a.
$endgroup$
– Varun Iyer
Sep 3 '14 at 1:28
add a comment |
$begingroup$
So I have my first quiz tomorrow and want to get off on the good foot, but I'm studying some problems and this one is particularly confusing...
There are a group of 191 students, of which 10 are taking French,
business and music; 36 are taking French and business; 20 are taking
French and music; 18 are taking business and music; 65 are taking
French; 76 are taking business; and 63 are taking music.
1a. How many are taking French or business (or both)? 1b. How many are
taking music or French (or both) but not business?
It might just be all the numbers confusing me, but I'm kind of lost. For 1a: the answer I got was 177 but I don't think that can be right, as there are 63 taking music, which would push the total over 191...
I haven't even gotten through 1b, and I don't want to look at the answer just yet, as I'm trying to work my way through it. Could anyone provide a bit of help?
discrete-mathematics
asked Sep 3 '14 at 1:02
secondublysecondubly
2921418
$endgroup$
So I have my first quiz tomorrow and want to get off on the good foot, but I'm studying some problems and this one is particularly confusing...
There are a group of 191 students, of which 10 are taking French,
business and music; 36 are taking French and business; 20 are taking
French and music; 18 are taking business and music; 65 are taking
French; 76 are taking business; and 63 are taking music.
1a. How many are taking French or business (or both)? 1b. How many are
taking music or French (or both) but not business?
It might just be all the numbers confusing me, but I'm kind of lost. For 1a: the answer I got was 177 but I don't think that can be right, as there are 63 taking music, which would push the total over 191...
I haven't even gotten through 1b, and I don't want to look at the answer just yet, as I'm trying to work my way through it. Could anyone provide a bit of help?
discrete-mathematics
discrete-mathematics
asked Sep 3 '14 at 1:02
secondublysecondubly
2921418
asked Sep 3 '14 at 1:02
secondublysecondubly
2921418
asked Sep 3 '14 at 1:02
secondublysecondubly
2921418
asked Sep 3 '14 at 1:02
secondublysecondubly
2921418
asked Sep 3 '14 at 1:02
secondublysecondubly
2921418
2921418
$begingroup$
Draw a venn diagram
$endgroup$
– voldemort
Sep 3 '14 at 1:03
$begingroup$
I don't think you were correct, please look at my answer for clarification. The number of music students doesn't affect the value in 1a.
$endgroup$
– Varun Iyer
Sep 3 '14 at 1:28
add a comment |
$begingroup$
Draw a venn diagram
$endgroup$
– voldemort
Sep 3 '14 at 1:03
$begingroup$
I don't think you were correct, please look at my answer for clarification. The number of music students doesn't affect the value in 1a.
$endgroup$
– Varun Iyer
Sep 3 '14 at 1:28
$begingroup$
Draw a venn diagram
$endgroup$
– voldemort
Sep 3 '14 at 1:03
$begingroup$
Draw a venn diagram
$endgroup$
– voldemort
Sep 3 '14 at 1:03
$begingroup$
I don't think you were correct, please look at my answer for clarification. The number of music students doesn't affect the value in 1a.
$endgroup$
– Varun Iyer
Sep 3 '14 at 1:28
$begingroup$
I don't think you were correct, please look at my answer for clarification. The number of music students doesn't affect the value in 1a.
$endgroup$
– Varun Iyer
Sep 3 '14 at 1:28
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is a typical Principle of Inclusion Exclusion principle problem. This principle allows one to count various unions and intersections when there are multiple not necessarily disjoint sets in a problem.
The most simple example is in the case of two sets; let A denote the set of positive multiples of 3 under 1000 (so 3,6,9,...,999), and let B denote the set of positive multiples of 5 under 1000 (so 5,10,15...,995). Then if |S| denotes the number of elements in a set S, the principle of inclusion and exclusion for A and B says:
$$
|A| + |B| - |A cap B| = |A cup B|
$$
There, $|A cap B|$ represents the multiples of 5 and 3 (so multiples of 15), while $|A cup B|$ represents the multiples of 5 OR 3. Understand the distinction.
In the case of three sets, as in your problem (F for students in French, B for students in business, and M for students in music), the principle says:
$$
|F| + |B| + |M| - |F cap B| - |F cap M| - |B cap M| + |F cap B cap M| = |F cup B cup M|
$$
In the equation above, $F cup B$ (for example) represents the students in both French and business, whereas $F cup B cup M$ represents the set of all students (so 191). You can use this expression to solve your problem!
answered Sep 3 '14 at 1:20
Likith GovindaiahLikith Govindaiah
361
$endgroup$
add a comment |
$begingroup$
You can either draw a venn diagram (if you are more of a visual learner), or use the Principle of Inclusion-Exclusion.
So we have three probable events $$A, B, C$$
And from the question we have that:
$$P(A cap B cap C) = 10 $$
In order to find the number of students taking only French, we can say that:
$$P(A cup B) = P(A) + P(B) - P(A cap B)$$
$$= 65 + 76 - 36 = 105$$
Can you use this idea to find the answer in part 1 b.?
answered Sep 3 '14 at 1:28
Varun IyerVarun Iyer
5,317826
$endgroup$
$begingroup$
Wow, I wish I this concept was actually taught in this section (they teach us about associate/distributive algebra of sets, but not this particular principle, which I'm guessing comes later). I drew a Venn Diagram and was absolutely lost - but this makes it so much easier!
$endgroup$
– secondubly
Sep 3 '14 at 1:34
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is a typical Principle of Inclusion Exclusion principle problem. This principle allows one to count various unions and intersections when there are multiple not necessarily disjoint sets in a problem.
The most simple example is in the case of two sets; let A denote the set of positive multiples of 3 under 1000 (so 3,6,9,...,999), and let B denote the set of positive multiples of 5 under 1000 (so 5,10,15...,995). Then if |S| denotes the number of elements in a set S, the principle of inclusion and exclusion for A and B says:
$$
|A| + |B| - |A cap B| = |A cup B|
$$
There, $|A cap B|$ represents the multiples of 5 and 3 (so multiples of 15), while $|A cup B|$ represents the multiples of 5 OR 3. Understand the distinction.
In the case of three sets, as in your problem (F for students in French, B for students in business, and M for students in music), the principle says:
$$
|F| + |B| + |M| - |F cap B| - |F cap M| - |B cap M| + |F cap B cap M| = |F cup B cup M|
$$
In the equation above, $F cup B$ (for example) represents the students in both French and business, whereas $F cup B cup M$ represents the set of all students (so 191). You can use this expression to solve your problem!
answered Sep 3 '14 at 1:20
Likith GovindaiahLikith Govindaiah
361
$endgroup$
add a comment |
$begingroup$
This is a typical Principle of Inclusion Exclusion principle problem. This principle allows one to count various unions and intersections when there are multiple not necessarily disjoint sets in a problem.
The most simple example is in the case of two sets; let A denote the set of positive multiples of 3 under 1000 (so 3,6,9,...,999), and let B denote the set of positive multiples of 5 under 1000 (so 5,10,15...,995). Then if |S| denotes the number of elements in a set S, the principle of inclusion and exclusion for A and B says:
$$
|A| + |B| - |A cap B| = |A cup B|
$$
There, $|A cap B|$ represents the multiples of 5 and 3 (so multiples of 15), while $|A cup B|$ represents the multiples of 5 OR 3. Understand the distinction.
In the case of three sets, as in your problem (F for students in French, B for students in business, and M for students in music), the principle says:
$$
|F| + |B| + |M| - |F cap B| - |F cap M| - |B cap M| + |F cap B cap M| = |F cup B cup M|
$$
In the equation above, $F cup B$ (for example) represents the students in both French and business, whereas $F cup B cup M$ represents the set of all students (so 191). You can use this expression to solve your problem!
answered Sep 3 '14 at 1:20
Likith GovindaiahLikith Govindaiah
361
$endgroup$
add a comment |
$begingroup$
This is a typical Principle of Inclusion Exclusion principle problem. This principle allows one to count various unions and intersections when there are multiple not necessarily disjoint sets in a problem.
The most simple example is in the case of two sets; let A denote the set of positive multiples of 3 under 1000 (so 3,6,9,...,999), and let B denote the set of positive multiples of 5 under 1000 (so 5,10,15...,995). Then if |S| denotes the number of elements in a set S, the principle of inclusion and exclusion for A and B says:
$$
|A| + |B| - |A cap B| = |A cup B|
$$
There, $|A cap B|$ represents the multiples of 5 and 3 (so multiples of 15), while $|A cup B|$ represents the multiples of 5 OR 3. Understand the distinction.
In the case of three sets, as in your problem (F for students in French, B for students in business, and M for students in music), the principle says:
$$
|F| + |B| + |M| - |F cap B| - |F cap M| - |B cap M| + |F cap B cap M| = |F cup B cup M|
$$
In the equation above, $F cup B$ (for example) represents the students in both French and business, whereas $F cup B cup M$ represents the set of all students (so 191). You can use this expression to solve your problem!
answered Sep 3 '14 at 1:20
Likith GovindaiahLikith Govindaiah
361
$endgroup$
This is a typical Principle of Inclusion Exclusion principle problem. This principle allows one to count various unions and intersections when there are multiple not necessarily disjoint sets in a problem.
The most simple example is in the case of two sets; let A denote the set of positive multiples of 3 under 1000 (so 3,6,9,...,999), and let B denote the set of positive multiples of 5 under 1000 (so 5,10,15...,995). Then if |S| denotes the number of elements in a set S, the principle of inclusion and exclusion for A and B says:
$$
|A| + |B| - |A cap B| = |A cup B|
$$
There, $|A cap B|$ represents the multiples of 5 and 3 (so multiples of 15), while $|A cup B|$ represents the multiples of 5 OR 3. Understand the distinction.
In the case of three sets, as in your problem (F for students in French, B for students in business, and M for students in music), the principle says:
$$
|F| + |B| + |M| - |F cap B| - |F cap M| - |B cap M| + |F cap B cap M| = |F cup B cup M|
$$
In the equation above, $F cup B$ (for example) represents the students in both French and business, whereas $F cup B cup M$ represents the set of all students (so 191). You can use this expression to solve your problem!
answered Sep 3 '14 at 1:20
Likith GovindaiahLikith Govindaiah
361
answered Sep 3 '14 at 1:20
Likith GovindaiahLikith Govindaiah
361
answered Sep 3 '14 at 1:20
Likith GovindaiahLikith Govindaiah
361
answered Sep 3 '14 at 1:20
Likith GovindaiahLikith Govindaiah
361
361
add a comment |
add a comment |
$begingroup$
You can either draw a venn diagram (if you are more of a visual learner), or use the Principle of Inclusion-Exclusion.
So we have three probable events $$A, B, C$$
And from the question we have that:
$$P(A cap B cap C) = 10 $$
In order to find the number of students taking only French, we can say that:
$$P(A cup B) = P(A) + P(B) - P(A cap B)$$
$$= 65 + 76 - 36 = 105$$
Can you use this idea to find the answer in part 1 b.?
answered Sep 3 '14 at 1:28
Varun IyerVarun Iyer
5,317826
$endgroup$
$begingroup$
Wow, I wish I this concept was actually taught in this section (they teach us about associate/distributive algebra of sets, but not this particular principle, which I'm guessing comes later). I drew a Venn Diagram and was absolutely lost - but this makes it so much easier!
$endgroup$
– secondubly
Sep 3 '14 at 1:34
add a comment |
$begingroup$
You can either draw a venn diagram (if you are more of a visual learner), or use the Principle of Inclusion-Exclusion.
So we have three probable events $$A, B, C$$
And from the question we have that:
$$P(A cap B cap C) = 10 $$
In order to find the number of students taking only French, we can say that:
$$P(A cup B) = P(A) + P(B) - P(A cap B)$$
$$= 65 + 76 - 36 = 105$$
Can you use this idea to find the answer in part 1 b.?
answered Sep 3 '14 at 1:28
Varun IyerVarun Iyer
5,317826
$endgroup$
$begingroup$
Wow, I wish I this concept was actually taught in this section (they teach us about associate/distributive algebra of sets, but not this particular principle, which I'm guessing comes later). I drew a Venn Diagram and was absolutely lost - but this makes it so much easier!
$endgroup$
– secondubly
Sep 3 '14 at 1:34
add a comment |
$begingroup$
You can either draw a venn diagram (if you are more of a visual learner), or use the Principle of Inclusion-Exclusion.
So we have three probable events $$A, B, C$$
And from the question we have that:
$$P(A cap B cap C) = 10 $$
In order to find the number of students taking only French, we can say that:
$$P(A cup B) = P(A) + P(B) - P(A cap B)$$
$$= 65 + 76 - 36 = 105$$
Can you use this idea to find the answer in part 1 b.?
answered Sep 3 '14 at 1:28
Varun IyerVarun Iyer
5,317826
$endgroup$
You can either draw a venn diagram (if you are more of a visual learner), or use the Principle of Inclusion-Exclusion.
So we have three probable events $$A, B, C$$
And from the question we have that:
$$P(A cap B cap C) = 10 $$
In order to find the number of students taking only French, we can say that:
$$P(A cup B) = P(A) + P(B) - P(A cap B)$$
$$= 65 + 76 - 36 = 105$$
Can you use this idea to find the answer in part 1 b.?
answered Sep 3 '14 at 1:28
Varun IyerVarun Iyer
5,317826
answered Sep 3 '14 at 1:28
Varun IyerVarun Iyer
5,317826
answered Sep 3 '14 at 1:28
Varun IyerVarun Iyer
5,317826
answered Sep 3 '14 at 1:28
Varun IyerVarun Iyer
5,317826
5,317826
$begingroup$
Wow, I wish I this concept was actually taught in this section (they teach us about associate/distributive algebra of sets, but not this particular principle, which I'm guessing comes later). I drew a Venn Diagram and was absolutely lost - but this makes it so much easier!
$endgroup$
– secondubly
Sep 3 '14 at 1:34
add a comment |
$begingroup$
Wow, I wish I this concept was actually taught in this section (they teach us about associate/distributive algebra of sets, but not this particular principle, which I'm guessing comes later). I drew a Venn Diagram and was absolutely lost - but this makes it so much easier!
$endgroup$
– secondubly
Sep 3 '14 at 1:34
$begingroup$
Wow, I wish I this concept was actually taught in this section (they teach us about associate/distributive algebra of sets, but not this particular principle, which I'm guessing comes later). I drew a Venn Diagram and was absolutely lost - but this makes it so much easier!
$endgroup$
– secondubly
Sep 3 '14 at 1:34
$begingroup$
Wow, I wish I this concept was actually taught in this section (they teach us about associate/distributive algebra of sets, but not this particular principle, which I'm guessing comes later). I drew a Venn Diagram and was absolutely lost - but this makes it so much easier!
$endgroup$
– secondubly
Sep 3 '14 at 1:34
add a comment |
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$begingroup$
Draw a venn diagram
$endgroup$
– voldemort
Sep 3 '14 at 1:03
$begingroup$
I don't think you were correct, please look at my answer for clarification. The number of music students doesn't affect the value in 1a.
$endgroup$
– Varun Iyer
Sep 3 '14 at 1:28