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Question: The set \(\left( A\cup B\cup C \right)\cap \left( A\cap B'\cap C' \right)'\cap C'\) is equal to \[...

The set (ABC)(ABC)C\left( A\cup B\cup C \right)\cap \left( A\cap B'\cap C' \right)'\cap C' is equal to

& A.B\cap C' \\\ & B.A\cap C \\\ & C.B\cap C' \\\ & D.A\cap C' \\\ \end{aligned}$$
Explanation

Solution

In solving these type of questions, we will use various properties of set to reach the final simplified answer. Properties which we will use are:
(i) (AB)=AB\left( A\cup B \right)'=A'\cap B' which is called De Morgan's first law.
(ii) (AB)=AB\left( A\cap B \right)'=A'\cup B' which is called De Morgan's second law.
(iii) (AB)C=A(BC)\left( A\cup B \right)\cup C=A\cup \left( B\cup C \right) which is called association law.
(iv) (AB)(AC)=A(BC)\left( A\cup B \right)\cap \left( A\cup C \right)=A\cup \left( B\cap C \right) which is called distributive law.
(v) (AB)=(BA)\left( A\cup B \right)=\left( B\cup A \right) which is called commutative law.
(vi) A=AA\cup \varnothing =A which is called law of identity element.
(vii) AA=A\cap A'=\varnothing which is called complement property.

Complete step-by-step answer:
We are given set (ABC)(ABC)C\left( A\cup B\cup C \right)\cap \left( A\cap B'\cap C' \right)'\cap C'
Using distributive law, we will solve first two terms together and then third term.
[(ABC)(ABC)]C\left[ \left( A\cup B\cup C \right)\cap \left( A\cap B'\cap C' \right)' \right]\cap C'
We will use distributive law in both terms to combine BC and BCB\cup C\text{ and }B'\cap C' we get:
[(A(BC))(A(BC))]C\left[ \left( A\cup \left( B\cup C \right) \right)\cap \left( A\cap \left( B'\cap C' \right) \right)' \right]\cap C'
Let us use De Morgan law in second term, so we get:
[(A(BC))(A(BC))]C\left[ \left( A\cup \left( B\cup C \right) \right)\cap \left( A'\cap \left( B\cap C \right) \right) \right]\cap C'
Using commutative rule, we get:
[((BC)A)((BC)A)]C\left[ \left( \left( B\cup C \right)\cup A \right)\cap \left( \left( B\cup C \right)\cup A' \right) \right]\cap C'
Using distributive law, we get:
[(BC)(AA)]C\left[ \left( B\cup C \right)\cup \left( A\cap A' \right) \right]\cap C'
As we know, AA=A\cap A'=\varnothing using complement property, therefore, equation becomes
[(BC)]C\left[ \left( B\cup C \right)\cup \varnothing \right]\cap C'
As we know, A=AA\cup \varnothing =A using law of identity element, we get:
[BC]C\left[ B\cup C \right]\cap C'
Using distributive law we get:
(BC)(CC)\left( B\cap C' \right)\cup \left( C\cap C' \right)
As we know CC=C\cap C'=\varnothing using complement property, therefore, equation becomes
(BC)\left( B\cap C' \right)\cup \varnothing
Since, A=AA\cup \varnothing =A using law of identity element we get:
(BC)=BC\left( B\cap C' \right)\cup \varnothing =B\cap C'
Hence, (ABC)(ABC)C\left( A\cup B\cup C \right)\cap \left( A\cap B'\cap C' \right)'\cap C' is equal to BCB\cap C'

So, the correct answer is “Option C”.

Note: Students should remember all the properties of sets. Notation ‘C^{C}’ can also be used for representing the complement of a set instead of ‘'’ only.
Students should not get confused with ,\cup ,\cap as both are completely different signs.