In a class of 30 pupils,12 take needle work,1 | vedclass

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(A) Let $N$,$P$,and $H$ represent the sets of students taking needle work,physics,and history,respectively.Given: $n(N) = 12$,$n(P) = 16$,$n(H) = 18$,

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$16$

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(A) Let $N$,$P$,and $H$ represent the sets of students taking needle work,physics,and history,respectively.Given: $n(N) = 12$,$n(P) = 16$,$n(H) = 18$,and $n(N \cup P \cup H) = 30$.Since no one takes all three subjects,$n(N \cap P \cap H) = 0$.Using the principle of inclusion-exclusion for three sets:$n(N \cup P \cup H) = n(N) + n(P) + n(H) - [n(N \cap P) + n(P \cap H) + n(N \cap H)] + n(N \cap P \cap H)$.Substituting the values:$30 = 12 + 16 + 18 - [n(N \cap P) + n(P \cap H) + n(N \cap H)] + 0$.$30 = 46 - [n(N \cap P) + n(P \cap H) + n(N \cap H)]$.Therefore,$n(N \cap P) + n(P \cap H) + n(N \cap H) = 46 - 30 = 16$.The number of students taking exactly two subjects is given by $[n(N \cap P) + n(P \cap H) + n(N \cap H)] - 3n(N \cap P \cap H)$.Since $n(N \cap P \cap H) = 0$,the number of students taking exactly two subjects is $16 - 3(0) = 16$.

In a class of $30$ pupils,$12$ take needle work,$16$ take physics,and $18$ take history. If all the $30$ students take at least one subject and no one takes all three,then the number of pupils taking $2$ subjects is:

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