Find the number of elements in A 1 ∪ A 2 ∪ A 3 if there are 200 elements in A 1 , 1000 in A 2 , and 5, 000 in A 3 if (a) A 1 ⊆ A 2 and A 2 ⊆ A 3 . (b) the sets are pairwise disjoint. (c) There are two elements in common to each pair of sets and one element in all three sets.

Question

09-12-2019
Mathematics

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Find the number of elements in A 1 ∪ A 2 ∪ A 3 if there are 200 elements in A 1 , 1000 in A 2 , and 5, 000 in A 3 if (a) A 1 ⊆ A 2 and A 2 ⊆ A 3 . (b) the sets are pairwise disjoint. (c) There are two elements in common to each pair of sets and one element in all three sets.

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steffaniasierrag steffaniasierrag · Answer 1 · 2019-12-10T05:20:52+01:00

Answer:

a. 4600

b. 6200

c. 6193

Step-by-step explanation:

Let [tex]n(A)[/tex] the number of elements in A.

Remember, the number of elements in [tex]A_1 \cup A_2 \cup A_3[/tex] satisfies

[tex]n(A_1 \cup A_2 \cup A_3)=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)[/tex]

Then,

a) If [tex]A_1\subseteq A_2, n(A_1 \cap A_2)=n(A_1)=200[/tex], and if [tex]A_2\subseteq A_3, n(A_2\cap A_3)=n(A_2)=1000[/tex]

Since [tex]A_1\subseteq A_2\; and \; A_2\subseteq A_3, \; then \; A_1\cap A_2 \cap A_3= A_1[/tex]

So

[tex]n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-200-200-1000-200=4600[/tex]

b) Since the sets are pairwise disjoint

[tex]n(A_1 \cup A_2 \cup A_3)=\\n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\200+1000+5000-0-0-0-0=6200[/tex]

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

[tex]n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-2-2-2-1=6193[/tex]