Venn diagram

[Malyster]

Could someone post how to solve a venn diagram?

The students studied

20 English

20 Math

15 History

12 French

2 Math and english

2 Math and history

2 Math, history and french

1 Math, English, history and french

How many students studied at least one subject

 

[Aion]

You should be able to solve this problem by using the inclusion-exclusion principle. I would not use Venn diagrams in this case, since I don't think it extends to 4 sets. Use algebra instead. Suppose we denote the sets [imath]E[/imath] for English, [imath]M[/imath]for Math., [imath]H[/imath] for history, [imath]F[/imath] for french. Then we need to expand and simplify the following expression

[math]\vert E \cup M \cup H \cup F \vert = ...[/math]

 

[Aion]

You should be able to solve this problem by using the inclusion-exclusion principle. I would not use Venn diagrams in this case, since I don't think it extends to 4 sets. Use algebra instead. Suppose we denote the sets [imath]E[/imath] for English, [imath]M[/imath]for Math., [imath]H[/imath] for history, [imath]F[/imath] for french. Then we need to expand and simplify the following expression

[math]\vert E \cup M \cup H \cup F \vert = ...[/math]

Since the expansion is quite complicated I've found a link to the correct expression. From this, it doesn't seem like there's enough information to solve the problem. For example, we need to know how many students studied english and history, but that's not on the list given above.

 

[Dr.Peterson]

Could someone post how to solve a venn diagram?

The students studied

20 English

20 Math

15 History

12 French

2 Math and english

2 Math and history

2 Math, history and french

1 Math, English, history and french

How many students studied at least one subject

It is possible to draw a Venn diagram for 4 sets, but it is awkward:



The Venn diagram can serve as a way to organize your thinking while you analyze your data. I typically start with the most restrictive information (intersection of all regions), and work up the list, filling in other regions by subtraction.

But this diagram necessarily has 16 regions, and you only have 8 facts, which will not be enough to find all parts. All you can hope is that you don't need that in order to answer the specific question they asked. Here is the start of an attempt (which I haven't tried to finish):



I doubt it can be reasonably solved this way; where did the problem come from, and do you have any reason to think it should be solvable?

 

[Aion]

The Highlander posted one possible solution to this problem in a private group with me and OP, which I thought was unfortunate. I was thinking about reposting it here as well, but maybe he'll come around and do this instead.

 

[Steven G]

EnFnMnH=1
MnHnF=1
MnH=2
MnE=2
M=20
E=15
F=12
H=15

Lets consider H. It is counted multiple times. It is counted 1 time in EnFnMnH, 1 time in MnHnF, two times in MnH and 15 times in H.
Because EnFnMnH=1 we reduce MnHnF by 1, reduce MnH by 1 and H by 1.
Now because of MnHnF=0 (we reduced it by 1) we reduce HnF and H by 0.
Because MnH=1, we reduce H by 1
So H=13
Do this for the other 3 letters.
Now add up ALL the numbers that remain and that will be your answer.