Explanation
$$n(U) =60 (Universal \; Set) $$
$$n(P \cap Q ) = 5$$
$$n(Q \cap R ) = 10$$
$$n(P ) = 20$$
$$n(Q)=23$$
$$n(P \cup Q ) = n(P) + n(Q)- n(P \cap Q ) = 20+23-5 = 38$$
$$n(R)=n(U)-n(P \cup Q ) +n(Q\cap R)= 60-38+10 = 32$$
$$n(P \cup R ) = n(P) + n(R)- n(P \cap R ) = 20+32-0 =52$$
A-B means everything in A except for anything in $$A \cap B$$
Pick an element x. Let $$ x\in(A-B)$$
$$\therefore x \in A \; but \; x \notin B$$
$$\therefore x\in A , x\in B'$$
$$x\in (A \cap B') = A- (A \cap B)$$
or, $$x\in (A- B)$$
Hence, $$A-B = (A \cap B') = A- (A \cap B)$$
$$n(P) = 20$$
$$n(M) = 17$$
$$n(M \cap P) = 5$$
$$n(other \; subjects)=10$$
$$n(M\cup P) =n(M)+n(P)-n(M\cap P)$$
$$n(M\cup P) =17+20-5 = 32$$
Total students $$= n(P\cup M)+n (other \; subjects )$$
$$32+10=42$$
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