Question

How can I calculate \[A \cup B \cup C\]?

Answer 1

I just see this:
\[(A\cup B)\cup C\]

Answer 2

\[ n( (A\cup B) \cup C)\]=?

Answer 3

I'm a little lost, Is it that correct?

Answer 4

\[=(n(A)+n(B) - n(A\cap B))\cup n(C)\]

Answer 5

I don't want to speak up on this because I've never studied this stuff in particular. I do want to see the answer though, so I'm posting.

Answer 6

Ok!

Answer 7

Mmm ok, let's see:

Answer 8

\[n((A\cup B)\cup C)=n(A\cup B)+n(C)-n((A\cup B)\cap C)\]

Answer 9

but \[(A\cup B)\cap C=A\cap B\cap C\]

Answer 10

A union B union C simply means to combine the three sets. The order in which you do it does not matter.

Answer 11

So:
\[n((A\cup B)\cup C)=n(A) + n(B) - n(A\cap B) - n(A\cap B \cap C)\]

Answer 12

@Mertsj Yeah But I just wanted to do it orderly.

Answer 13

For example: Let A = {1,2,3} B = {2,4,6} C = {8}
Then A U B U C = {1,2,3,4,6,8}

Answer 14

Yes, the associative propertie.

Answer 15

ok then A U B U C = (A U B) U C or (A U C) U B

Answer 16

I didn't know how to get the number of elements of it.

Answer 17

depends on the number of elements in the individual sets.

Answer 18

I wanted to know on many elements the set \[A\cap B\cap C\] has.

Answer 19

I'm thinking

Answer 20

I think the formula is wrong =(

Answer 21

yeah I missed the \[n(C)\]

Answer 22

\[n(A\cup B\cup C)=n(A) + n(B) + n(C) - n(A\cap B) - n(A\cap B \cap C) \]

Answer 23

I guess that make sense.

Answer 24

n() means number of elements?

Answer 25

@myininaya Yes, is the number of elements.

Answer 26

Do you think I'm right?

Answer 27

Or Is there an easier way to calculate that?

Answer 28

I think I tend to overcomplicate things.

Answer 29

can't think of a way to explain it, but its the principle of inclusion-exclusion
http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle

Answer 30

\[n((A \cup B) \cup C)=n(A \cup B )+n(C)-n((A \cup B) \cap C)\]
\[=n(A)+n(B)-n(A \cap B)+n(C)-n(A \cup (B \cap C))\]
\[=n(A)+n(B)-n(A \cap B) +n(C)-[n(A)+n(B \cap C)-n(A \cap (B \cap C))]\]
\[=n(A)+n(B)+n(C)-n(A)-n(B \cap C)-n(A \cap B)+n(A \cap B \cap C)\]
\[=n(B)+n(C)-n(A \cap B) -n(B \cap C)+n(A \cap B \cap C)\]
is this what you are looking for?

Answer 31

i think i made a mistake

Answer 32

when i was trying to derive the formula

Answer 33

cazil's link provides the correct formula to use

Answer 34

Yeah, I think my formula is wrong too.

Answer 35

I was trying to use the following thm (its about probability but I think it also works for the number of elements)...
If C_1 and C_2 are events in C, then
P(C_1 U C_2)=P(C_1)+P(C_2)-P(C_1 and C_2)

Answer 36

i tried to let A U B be event C_1 first and let C be event C_2

Answer 37

yeah it's the exact same principle for probability.

Answer 38

and plugged into the formula
then I used this formula again

Answer 39

So did you get it, No-data?

Answer 40

omg I know what I did wrong...

Answer 41

No, I don't see why

Answer 42

I was following myininaya derivation.

Answer 43

Basically it says to add the number of each set , subtract the number in each intersection and then add back the number in the intersection of all three sets.

Answer 44

\[( A \cup B) \cap C=(A \cup B) \cap (B \cup C)\]

Answer 45

n(AUBUC)=n(a)+n(B)+n(C)-n(A intersect B)-n(A intersect C)- n(B intersect C) + n(A intersect B intersect C)

Answer 46

thats one part i messed up on i think i'm going to right this down again

Answer 47

Yes, I messed up on that part too!

Answer 48

Thank you myininaya

Answer 49

\[n(A\cup B\cup C)\]
\[=n(A)+n(B)+n(C)-n(A \cap B) -n(B \cap C)-n(A\cap C)+n(A \cap B \cap C)\]

Answer 50

we know, we want to prove it though.

Answer 51

Yeah, satellite73 I didn't know the formula so I tried to derive it.

Answer 52

proof? venn diagrams will provide a proof i think
element-wise will be a pain

Answer 53

Yes is a pain!

Answer 54

idea is you need to subtract off the intersections, just as you do with
\[A\cup B\] but when you subtract all three you have subtracted of the interersection
\[A\cap B\cap C\] three times so you have to add it back

Answer 55

That was mentioned earlier but I don't think it's quite what we're looking for. I think no-data wants a more formal proof.

Answer 56

ok
i like proofs of these with venn diagrams, which to me count as a valid proof.
in any case i think it is time for a glass of champagne
enjoy the new year!

Answer 57

\[(A\cup B)\cap C \neq (A\cup B) \cap (B\cup C)\]

Answer 58

yes the signs are backwards

Answer 59

yeah, but I think i will enjoy this a lot more if we can figure out this haha.

Answer 60

\[(A\cup B)\cap C = (A\cap B) \cup (B\cap C)\]

Answer 61

are you sure? That seems to contradict the Venn I'm looking at.

Answer 62

\[(A\cup B)\cap C = (A\cap C)\cup (B\cap C)\]

Answer 63

Is some kind of distributive law...

Answer 64

darn right

Answer 65

stupid B

Answer 66

i meant the set B by the way

Answer 67

haha

Answer 68

Where are you guys finding the intersection symbol?

Answer 69

\cap

Answer 70

Finally!

Answer 71

(AUB) intersect C = A intersect C + B intersect C - A intersect B intersect C

Answer 72

ok okay what about this:

Pretend again we have events C_1 and C_2 in C

so we know we can write
\[C_1 \cup C_2=C_1 \cup (C^c_1 \cap C_2) \text{ and } C_2=(C_1 \cap C_2) \cup (C^c_1 \cap C_2)\]
Then \[n(C_1 \cup C_2)=n(C_1)+n(C^c_1 \cap C_2)\]
and
\[n(C_2)=n(C_1 \cap C_2)+n(C_1^c \cap C_2)\]
so we can solve the second equation here for \[n(C_1^c \cap C_2)\]
and plug into the first to get
\[n(C_1 \cup C_2)=n(C_1)+n(C_2)-n(C_1 \cap C_2)\]

Answer 73

Did you wrote that myin? \[(A\cup B)\cap C=A\cap C + B\cap C - A\cap B \cap C\]

Answer 74

i don't recall writing that

Answer 75

Ahh no, it was mMertsj

Answer 76

I wrote that No-Data

Answer 77

Ok

Answer 78

I didn't understood quite wel the last equations you posted Myininaya.

Answer 79

I'm goin to post what i have , but i't will take me some time to type it.

Answer 80

I don't think I follow that proof either,
how did you sub\[C_1\cup C_2\]for\[C_1\cup( C_1\cap C_2)\]???

Answer 81

what?

Answer 82

\[n((A\cup B) \cup C) = n(A\cup B) + n(C) - n((A\cup B)\cap C)\]

Answer 83

well you say they are equal, I don't see how.

Answer 84

i don't see how either i think i wrote something else not what you wrote unless i made a type-o

Answer 85

\[C_1 \cup C_2=C_1 \cup (C^c_1 \cap C_2) \text{ and } C_2=(C_1 \cap C_2) \cup (C^c_1 \cap C_2) \]

Answer 86

\[ n((A\cup B) \cup C) = n(A) + n(B) - n(A\cap B) - n((A\cup B)\cap C)\]

Answer 87

I don't really understand either of those statements with C...

Answer 88

\[(A\cup B)\cap C = (A\cap C)\cup(B\cap C)\] So.

Answer 89

\[C_1^c \cap C_2=C_2 \text{ is what i see} \]

Answer 90

Do you see this?

Answer 91

\[n((A\cup B)\cap C) = n((A\cap C)\cup(B\cap C))\]

Answer 92

the intersection of not C_1 and C_2 is C_2

Answer 93

You're almost there No-data. Now use:
\[n((A∪B)∩C)=n((A ∩C)\cup(B∩C))\]
\[=n(A∩C)+n(B∩C)-n((A∩C)∩(B∩C))\]

Answer 94

intersection and union are associative also