Calculating Element Count in Disjoint Volumes Using Venn Diagram

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SUMMARY

The discussion centers on calculating the number of elements in the union of three sets (A1, A2, A3) using Venn diagrams. Participants analyze three scenarios: (a) pairwise disjoint sets, (b) overlapping pairs, and (c) all three sets with specific overlaps. The correct calculation for scenario (c) involves recognizing that there are 10 elements in common for each pair and 4 elements common to all three sets, leading to a final count of 11172 elements in the union.

PREREQUISITES
  • Understanding of Venn diagrams and set theory
  • Basic knowledge of union and intersection of sets
  • Familiarity with element counting in overlapping sets
  • Ability to interpret mathematical problems involving multiple sets
NEXT STEPS
  • Study advanced Venn diagram techniques for complex set problems
  • Learn about the principle of inclusion-exclusion in set theory
  • Explore combinatorial counting methods for overlapping sets
  • Practice solving problems involving unions and intersections of multiple sets
USEFUL FOR

Students studying mathematics, educators teaching set theory, and anyone interested in combinatorial analysis and problem-solving with Venn diagrams.

Petrus
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Calculate the number of elements in A1 U A2 U A3 if there are 97 elements in A1, 997 elements in A2 and 10098 elements in A3 in each of the following situations:
(a) The amounts are pairwise disjoint, i.e.
c3939de84ef91f8e6d3d9088938ee01.png
, too
ea477786ed42b36a173e5f8918b58a1.png
.

(b)
0a0762b1e265e781e6409e400e5f951.png
.

(c) there are 10 elements in common for each pair of volumes and 4 elements common to all three sets.

(a)I get answer 997+97+10098
(b)10098
(c) is the part I think I do wrong, will attach a picture of my venn diagram.
for some reason I get problem to attaché the picture but i get a1= 73, a2= 973, A3=10074 so the answer is (that is wrong I think) 73+10+10+10+4+10074+973
 
Last edited:
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Petrus said:
Calculate the number of elements in A1 U A2 U A3 if there are 97 elements in A1, 997 elements in A2 and 10098 elements in A3 in each of the following situations:
(a) The amounts are pairwise disjoint, i.e.
c3939de84ef91f8e6d3d9088938ee01.png
, too
ea477786ed42b36a173e5f8918b58a1.png
.

(b)
0a0762b1e265e781e6409e400e5f951.png
.

(c) there are 10 elements in common for each pair of volumes and 4 elements common to all three sets.

(a)I get answer 997+97+10098
(b)10098
(c) is the part I think I do wrong, will attach a picture of my venn diagram.
for some reason I get problem to attaché the picture but i get a1= 73, a2= 973, A3=10074 so the answer is (that is wrong I think) 73+10+10+10+4+10074+973

Hi Petrus!

Well, your (a) and (b) are right.
... so I guess we'll be waiting for your venn diagram...
 
I like Serena said:
Hi Petrus!

Well, your (a) and (b) are right.
... so I guess we'll be waiting for your venn diagram...
Hello Serena,
I can't uppload my picture but I explained how it looks like
 
Petrus said:
Hello Serena,
I can't uppload my picture but I explained how it looks like

Okay. Let me upload then.
A and B overlap with 10 elements in common.
The 3 sets together have 4 elements in common.

View attachment 706

This doesn't quite seem to match what you have...
 

Attachments

  • venn.png
    venn.png
    2.6 KB · Views: 114
I like Serena said:
Okay. Let me upload then.
A and B overlap with 10 elements in common.
The 3 sets together have 4 elements in common.

View attachment 706

This doesn't quite seem to match what you have...
Will it also be 6 on that A and C that you have not painted?
 
Petrus said:
Will it also be 6 on that A and C that you have not painted?

Well, A and C would overlap in 10 elements and 4 of those elements are shared by all three.
So yes, it will also be 6 in the remaining overlap of A and C.
 
I like Serena said:
Well, A and C would overlap in 10 elements and 4 of those elements are shared by all three.
So yes, it will also be 6 in the remaining overlap of A and C.
basically I have missunderstand the problem:( I thought it was 4 on a and b and c(that is correct) BUT its 10 on a and b... Thanks Sirena for help!:) I get the answer that is 11172

edit: after reading this again I notice how bad I explained. basically it was suposted to be 4 on a and b, when I calculated i calculated with 10 :P Unfortently I am not a verry good explainer soo I hope this explain ok:)
 
Last edited:
Petrus said:
basically I have missunderstand the problem:( I thought it was 4 on a and b and c(that is correct) BUT its 10 on a and b... Thanks Sirena for help!:) I get the answer that is 11172

Hmm, I get a different answer.
How did you find yours?
 
I like Serena said:
Hmm, I get a different answer.
How did you find yours?
I get the rest 6 I can't uppload picture so I will just type what I get
10082+987+81+6+6+6+4=11172

97-6-4-6=81
997-6-4-4=987
10098-6-4-4=10082

Hope this give you a clue :P Thanks for the help as well Serena!:)
 
  • #10
Petrus said:
997-6-4-4=987

This one doesn't seem quite right.

- - - Updated - - -

Petrus said:
10098-6-4-4=10082

And this one has the right result, but a wrong number on the left hand side.
 
  • #11
I like Serena said:
This one doesn't seem quite right.

- - - Updated - - -
And this one has the right result, but a wrong number on the left hand side.

hmm? How can I get correct?:S (its a problem that you answer through internet and if you answer correct it says 'correct' and I got correct:S?)
 
  • #12
Petrus said:
I get the rest 6 I can't uppload picture so I will just type what I get
10082+987+81+6+6+6+4=11172

97-6-4-6=81
997-6-4-4=987
10098-6-4-4=10082

Hope this give you a clue :P Thanks for the help as well Serena!:)

Petrus said:
hmm? How can I get correct?:S (its a problem that you answer through internet and if you answer correct it says 'correct' and I got correct:S?)

They should be like this:
97-6-4-6=81
997-6-4-6=981
10098-6-4-6=10082
 
  • #13
I like Serena said:
They should be like this:
97-6-4-6=81
997-6-4-6=981
10098-6-4-6=10082
That give me also correct?! Are you serious? This is bad that I can get correct from both...? I will talk to my teacher about this because its not good if I get correct from miss calculated + doing a wrong step
 
  • #14
Petrus said:
I get the rest 6 I can't uppload picture so I will just type what I get
10082+987+81+6+6+6+4=11172

97-6-4-6=81
997-6-4-4=987
10098-6-4-4=10082

Hope this give you a clue :P Thanks for the help as well Serena!:)

Petrus said:
That give me also correct?! Are you serious? This is bad that I can get correct from both...? I will talk to my teacher about this because its not good if I get correct from miss calculated + doing a wrong step

I don't understand what you mean, but your end result should be:
10082+981+81+6+6+6+4=11166
 
  • #15
I like Serena said:
I don't understand what you mean, but your end result should be:
10082+981+81+6+6+6+4=11166
What I mean is that this is a problem I get from a program. We are supposed to write the answer and it says if its correct or wrong, I get correct from my 'wrong answer' and the correct answer. Thanks to you I know what I did wrong and the correct answer...
 

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