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

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(A) Let $N, P, H$ be 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

Vedclass · Accepted Answer

$16$

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(A) Let $N, P, H$ be 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,$n(N \cap P \cap H) = 0$.The formula for the union of three sets is:$n(N \cup P \cup H) = n(N) + n(P) + n(H) - [n(N \cap P) + n(P \cap H) + n(N \cap H)] + n(N \cap P \cap H)$Substituting the values:$30 = 12 + 16 + 18 - [n(N \cap P) + n(P \cap H) + n(N \cap H)] + 0$$30 = 46 - [n(N \cap P) + n(P \cap H) + n(N \cap H)]$$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)$$= 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 exactly $2$ subjects is:

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