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(A) Let $n(M)=20$,$n(P)=25$,$n(C)=16$,and $n(M \cup P \cup C)=40$. Let $x$ be the number of students who passed in all three subjects.Using the Princi

Vedclass · Accepted Answer

$10$

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(A) Let $n(M)=20$,$n(P)=25$,$n(C)=16$,and $n(M \cup P \cup C)=40$. Let $x$ be the number of students who passed in all three subjects.Using the Principle of Inclusion-Exclusion:$n(M \cup P \cup C) = n(M) + n(P) + n(C) - [n(M \cap P) + n(P \cap C) + n(M \cap C)] + n(M \cap P \cap C)$$40 = 20 + 25 + 16 - [n(M \cap P) + n(P \cap C) + n(M \cap C)] + x$$40 = 61 - [n(M \cap P) + n(P \cap C) + n(M \cap C)] + x$$[n(M \cap P) + n(P \cap C) + n(M \cap C)] = 21 + x$We are given that $n(M \cap P) \leq 11$,$n(P \cap C) \leq 15$,and $n(M \cap C) \leq 15$.Summing these inequalities: $n(M \cap P) + n(P \cap C) + n(M \cap C) \leq 11 + 15 + 15 = 41$.Substituting the expression from the inclusion-exclusion principle:$21 + x \leq 41 \Rightarrow x \leq 20$.Also,$x$ must be less than or equal to the intersection of any two sets,so $x \leq 11$. Checking the constraints with the Venn diagram logic,the maximum value satisfying all conditions is $x = 10$.

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