Question 1047210: Let S be the universal set, where: S={1,2,3,...,18,19,20} Let sets A and B be subsets of S, where:?
A={2,5,6,7,14,18,20}
B={1,3,4,5,7,8,9,10,11,17,20}
C={1,2,3,4,6,10,12,14,15,16,17,20}
Find the number of elements in the set (A∩C) ∩ B n[(A∩C)∩Bc]
Found 2 solutions by stanbon, AnlytcPhil: Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Let S be the universal set, where: S={1,2,3,...,18,19,20} Let sets A and B be subsets of S, where:? {
A={2,5,6,7,14,18,20}
B={1,3,4,5,7,8,9,10,11,17,20}
C={1,2,3,4,6,10,12,14,15,16,17,20}
Find the number of elements in the set (A∩C) ∩ B n[(A∩C)∩Bc]
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(A and C) = n{2,6,14,20}
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(A and C) and B = {20}
B' = {2,6,12,13,14,15,16,18,19}
(A and C) and B' = {-}
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Ans:: Zero
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Cheers,
Stan H.
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Answer by AnlytcPhil(1807) (Show Source):
You can put this solution on YOUR website!
Find the number of elements in the set
(A∩C) ∩ B ∩ [(A∩C) ∩ Bc]
First we find
A∩C = {2,5,6,7,14,18,20} ∩ {1,2,3,4,6,10,12,14,15,16,17,20}
Take the elements in common to A and C:
A∩C = {2,6,14,20}
We find Bc = B={1,3,4,5,7,8,9,10,11,17,20}c
That's the set of all elements in the universal set S
which are not elements of B
S={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}
So
Bc = {2,6,12,13,14,15,16,18,19}
We find (A∩C) ∩ B
(A∩C) ∩ B = {2,6,14,20} ∩ {1,3,4,5,7,8,9,10,11,17,20}
By taking the elements in common:
(A∩C) ∩ B = {20} (There's only one!)
Next we find
[(A∩C) ∩ Bc] = {2,6,14,20} ∩ {2,6,12,13,14,15,16,18,19}
Taking the elements in common:
[(A∩C) ∩ Bc] = {2,6,14}
So
(A∩C) ∩ B ∩ [(A∩C) ∩ Bc]
{20} ∩ {2,6,14}
There are no elements in common, so the answer
is the empty set:
Answer = n(Ø) which is 0.
Edwin
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