Question

Find the number of elements in A1 ∪ A2 ∪ A3 if there are 100 elements in A1, 1000 in A2, and 10,000 in A3 if

a) A1 ⊆ A2 and A2 ⊆ A3.
b) the sets are pairwise disjoint.
c) there are two elements common to each pair of sets and one element in all three sets.

Answer 1

(a) 1000

(b) 11100

(c) 11095.

Step-by-step explanation:

(a) If A1 is a subset of A2 and A2 is a subset of A3, then all the elements of A1 are in A2 and all the elements of A2 are in A3.

Then, n(A1 n A2) = 100, n(A2 n A3) = 1000 , n(A1 n A3) = 100 and n(A1 n A2 n A3) = 100.

So, we get

`n(A1\cup A2\cup A3)\\\\=n(A1)+n(A2)+n(A3)-n(A1\cap A2)-n(A2\cap A3)-n(A1\cap A3)+n(A1\cap A2\cap A3)\\\\=100+1000+1000-100-1000-100+100\\\\=1000.`

(b) If the sets are pairwise disjoint, then

n(A1 n A2) = n(A2 n A3) = n(A1 n A3) = n(A1 n A2 n A3) = 0.

So, we get

`n(A1\cup A2\cup A3)\\\\=n(A1)+n(A2)+n(A3)\\\\=100+1000+10000\\\\=11100.`

(c) If there are two elements common to each pair of sets and one element in all three sets, then

n(A1 n A2) = 2, n(A2 n A3) = 2, n(A1 n A3) = 2 and n(A1 n A2 n A3) = 1.

So, we get

`n(A1\cup A2\cup A3)\\\\=n(A1)+n(A2)+n(A3)-n(A1\cap A2)-n(A2\cap A3)-n(A1\cap A3)-n(A1\cap A2\cap A3)\\\\=100+1000+1000-2-2-2+1\\\\=11100-5\\\\=11095.`