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Question: For any two sets \(S\) and \(T\), \(S\Delta T\) is defined as the set of all elements that belong to...

For any two sets SS and TT, SΔTS\Delta T is defined as the set of all elements that belong to either SS or TT but not both, that is SΔT=(ST)(ST)S\Delta T = \left( {S \cup T} \right) - \left( {S \cap T} \right). Let AA, BB and CC be sets such that ABC=ϕA \cap B \cap C = \phi , and the number of elements in each of AΔBA\Delta B, BΔCB\Delta C and CΔAC\Delta A equals to 100. Then the number of elements in ABCA \cup B \cup C equals

Explanation

Solution

We will use the formula, n(ABC)=n(A)+n(B)+n(C)n(AB)n(BC)n(CA)+n(ABC)n\left( {A \cup B \cup C} \right) = n\left( A \right) + n\left( B \right) + n\left( C \right) - n\left( {A \cap B} \right) - n\left( {B \cap C} \right) - n\left( {C \cap A} \right) + n\left( {A \cap B \cap C} \right) to find the value of number of elements ofABCA \cup B \cup C. We will then substitute the value of ABC=ϕA \cap B \cap C = \phi . We will then use the Venn diagram to find the value of the unknowns.

Complete step-by-step answer:
We have to find the value of ABCA \cup B \cup C
We know that n(ABC)=n(A)+n(B)+n(C)n(AB)n(BC)n(CA)+n(ABC)n\left( {A \cup B \cup C} \right) = n\left( A \right) + n\left( B \right) + n\left( C \right) - n\left( {A \cap B} \right) - n\left( {B \cap C} \right) - n\left( {C \cap A} \right) + n\left( {A \cap B \cap C} \right)
Also, we are given that ABC=ϕA \cap B \cap C = \phi and n(ABC)=0n\left( {A \cap B \cap C} \right) = 0
n(ABC)=n(A)+n(B)+n(C)n(AB)n(BC)n(CA)n\left( {A \cup B \cup C} \right) = n\left( A \right) + n\left( B \right) + n\left( C \right) - n\left( {A \cap B} \right) - n\left( {B \cap C} \right) - n\left( {C \cap A} \right) eqn. (1)
We are also given that
n(AΔB)=100 n((AB)(AB))=100  n\left( {A\Delta B} \right) = 100 \\\ \Rightarrow n\left( {\left( {A \cup B} \right) - \left( {A \cap B} \right)} \right) = 100 \\\
Let us now simplify the expression (AB)(AB)\left( {A \cup B} \right) - \left( {A \cap B} \right) using Venn-Diagram.
If AA and BB are two sets, then the intersection includes only the common portion and the union includes all the elements of both the sets.

Hence, n((AB)(AB))=n(A)+n(B)n(AB)n\left( {\left( {A \cup B} \right) - \left( {A \cap B} \right)} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)
Similarly, we can write this for any two sets.
Therefore, we can write equation (1) as
n(ABC)=n(A)+n(B)+n(C)n(AB)n(BC)n(CA) n(ABC)=(n(A)+n(B)n(AB))+n(C)n(BC)n(CA)  n\left( {A \cup B \cup C} \right) = n\left( A \right) + n\left( B \right) + n\left( C \right) - n\left( {A \cap B} \right) - n\left( {B \cap C} \right) - n\left( {C \cap A} \right) \\\ \Rightarrow n\left( {A \cup B \cup C} \right) = \left( {n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)} \right) + n\left( C \right) - n\left( {B \cap C} \right) - n\left( {C \cap A} \right) \\\
Here, we do not have any condition on the number of intersections of two sets, therefore, ABCA \cup B \cup C cannot be determined.

Note: We use Venn diagrams to represent the sets. ABA \cup B is the set of elements from both the sets AA and BB, similarly, ABA \cap B has all common elements of AA and BB. The set ABA - B has the elements of set AA which are not a part of ABA \cap B.