To prove, that \(A-(B\cap C)=(A-B)\cup (A-C).\)
Let  \(x\in A-(B\cap C)\)
\(\Rightarrow\quad x\in A\quad and\quad x\notin (B\cap C)\)
\(\Rightarrow \quad x\in A\quad and\quad (x\notin B\quad or\quad x \notin C)\)
\(\Rightarrow \quad (x\in A\quad and\quad x\notin B)\quad or\quad (x\in A\quad and\quad (x\notin C)\)
\(\Rightarrow \quad x\in (A-B)\quad or\quad x\in (A-c)\)
\(\Rightarrow \quad x\in (A-B)\cup (A-C)\)
\(\because \quad A-(B\cap C)\subseteq (A-B)\cup (A-C)\quad \quad \quad .....(i)\)
Again, let y \(\in\) \((A-B)\cup (A-C)\)
\(\Rightarrow \quad y\in (A-B)\quad or\quad y\in (A-c)\)
\(\Rightarrow \quad (y\in A\quad and\quad y\notin B)\quad or(y\in A\quad and\quad y\notin C)\)
\(\Rightarrow \quad y\in A\quad and\quad (y\notin B\quad or\quad y\notin C)\)
\(\Rightarrow\quad y\in A\quad and\quad y\notin (B\cap C)\Rightarrow y\epsilon A-(B\cap C)\Rightarrow y\epsilon A-(B\cap C)\)
\(\Rightarrow \quad (A-B)\cup (A-C)\subseteq A-(B\cap C) \quad \quad ........(ii)\)
From Eqs. (i) and (ii), we get
\(A-(B\cap C)=(A-B)\cup (A-C).\)

\(n(E\cap S\cap \overline { E } )=n(E\cap S)-n(E\cap M\cap S)\)
=74=3

n(E)=15,n(M)=12,n(s)=8,\(n(E\cap M)=6\)
\(n(M\cap S)=7,n(E\cap S)=4,n(E\cap M\cap S)=4\)
(i)\(n(E\cap M\cap \overline { S } )=n(E\cap M)-n(E\cap M\cap S)\)
=6-4=2

The number of people who were liking neither (A) 'Aitraaz' nor 'Veer Zaara' (V) is given by
\(n({ V }^{ ' }\cap { A }^{ , }) ={ n(V\cup A) }^{ ' }\)
=\(n(\cup )-n(V\cup A)\)
=\(n(\cup )-n(V)-n(A)+n(V\cap A)\)