Question

Let S be the universal set, where: S = { 1 , 2 , 3 , ... , 18 , 19 , 20 } Let sets A , B , and C be subsets of S , where: Set A = { 1 , 6 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 18 , 19 , 20 } Set B = { 1 , 3 , 4 , 6 , 8 , 9 , 10 , 11 , 12 , 15 , 20 } Set C = { 3 , 6 , 7 , 8 , 9 , 10 , 11 , 13 , 14 , 17 } Find the number of elements in the set ( A ∪ B ∪ C ) n ( A ∪ B ∪ C ) = Find the number of elements in the set ( A ∩ B ∩ C ) n ( A ∩ B ∩ C ) =

Answer 1

The number of elements in the set ( A ∪ B ∪ C ) ∩ ( A ∪ B ∪ C ) is 20 and ( A ∩ B ∩ C ) n ( A ∩ B ∩ C ) is 2 .

Step-by-step explanation:

Part 1: ( A ∪ B ∪ C ) n ( A ∪ B ∪ C )

Since (A ∪ B ∪ C) is already the entire universal set S (as all elements in each set are included), taking the union again with itself doesn't change anything. Therefore, ( A ∪ B ∪ C ) n ( A ∪ B ∪ C ) = S. The number of elements in S is 20.

Part 2: ( A ∩ B ∩ C ) n ( A ∩ B ∩ C )

Finding the intersection of three sets can be a bit trickier. However, since we're taking the intersection of the same set with itself, it simply means finding the elements that are common in all three sets.

Looking at the given sets:

A ∩ B = { 1, 6, 10, 11, 12, 15, 20 }

B ∩ C = { 3, 6, 8, 9, 10, 11 }

A ∩ C = { 6, 11, 13, 14 }

Therefore, ( A ∩ B ∩ C ) = { 6, 11 }. Taking the intersection of this set with itself again simply yields the same result. So, ( A ∩ B ∩ C ) n ( A ∩ B ∩ C ) = { 6, 11 }. Therefore, there are 2 elements in this set.

Answer 2

1. A B C=(6,10,11)

2.A B C =(1,3,4,5,7,8,9,12,13,14......20)