Calculating Element Count in Disjoint Volumes Using Venn Diagram

[Petrus]

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.

, too

.

(b)

.

(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

 

[I like Serena]

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.

, too

.

(b)

.

(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...

 

[Petrus]

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

 

[I like Serena]

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...

 

[Petrus]

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?

 

[I like Serena]

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.

 

[Petrus]

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.

Basicly 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. Basicly 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:)

 

[I like Serena]

Basicly 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?

 

[Petrus]

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!:)

 

[I like Serena]

997-6-4-4=987

This one doesn't seem quite right.

- - - Updated - - -

10098-6-4-4=10082

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

 

[Petrus]

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?)

 

[I like Serena]

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!:)

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

 

[Petrus]

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

 

[I like Serena]

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!:)

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

 

[Petrus]

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...