Question

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.

Answer 1

Answer:

a. 4600

b. 6200

c. 6193

Step-by-step explanation:

Let $n(A)$ the number of elements in A.

Remember, the number of elements in $A_1 \cup A_2 \cup A_3$ satisfies

$n(A_1 \cup A_2 \cup A_3)=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)$

Then,

a) If $A_1\subseteq A_2, n(A_1 \cap A_2)=n(A_1)=200$, and if $A_2\subseteq A_3, n(A_2\cap A_3)=n(A_2)=1000$

Since $A_1\subseteq A_2\; and \; A_2\subseteq A_3, \; then \; A_1\cap A_2 \cap A_3= A_1$

So

$n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-200-200-1000-200=4600$

b) Since the sets are pairwise disjoint

$n(A_1 \cup A_2 \cup A_3)=\\n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\200+1000+5000-0-0-0-0=6200$

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

$n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-2-2-2-1=6193$