If in a class of $100$ students, $60$ like mathematics, $72$ like physics, $68$ like chemistry and no student likes all three subjects, then the number of students who didn't like mathematics and chemistry is ?

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Solution

$n (M \cup P \cup C) = n (M) + n (P) + n (C) - n (M \cap P) - n (M \cap C) - n (P \cap C)$
$100 = 60 + 72 + 68 - (x + y + z) + 0 \Rightarrow x + y + z = 100$

So no student like only maths, or only physics or only chemistry

$\begin{matrix} x + y = 60 x + z = 72 y + z = 68 \end{matrix} ⎫ ⎪ ⎬ ⎪ ⎭ \Rightarrow y = 28$

Now,
$n (M^{'} \cap C^{'}) = 100 - n (M \cup C)$ $= 100 - (60 + 68 - y) = 28 - 28 = 0$

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If in a class of 100 students, 60 like mathematics, 72 like physics, 68 like chemistry and no student likes all three subjects, then the number of students who didn't like mathematics and chemistry is ?

n(M∪P∪C)=n(M)+n(P)+n(C)−n(M∩P)−n(M∩C)−n(P∩C) 100=60+72+68−(x+y+z)+0⇒x+y+z=100 So no student like only maths, or only physics or only chemistry x+y=60x+z=72y+z=68⎫⎪⎬⎪⎭⇒y=28 Now, n(M′∩C′)=100−n(M∪C)=100−(60+68−y)=28−28=0

If in a class of $100$ students, $60$ like mathematics, $72$ like physics, $68$ like chemistry and no student likes all three subjects, then the number of students who didn't like mathematics and chemistry is ?