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

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Question

(a) Given $n(N) = 12$, $n(P) = 16$, $n(H) = 18$, $n(N \cup P \cup H) = 30$

From, $n(N \cup P \cup H) = n(N) + n(P) + n(H) - n(N \cap P)$

$ - n(P

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(a) Given $n(N) = 12$, $n(P) = 16$, $n(H) = 18$, $n(N \cup P \cup H) = 30$

From, $n(N \cup P \cup H) = n(N) + n(P) + n(H) - n(N \cap P)$

$ - n(P \cap H) - n(N \cap H) + n(N \cap P \cap H)$

 $	herefore n(N \cap P) + n(P \cap H) + n(N \cap H) = 16$

Now, number of pupils taking two subjects

$ = n(N \cap P) + n(P \cap H) + n(N \cap H) - 3n(N \cap P \cap H)$

$ = 16 - 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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