In a class of 140 students numbered 1 to 140, | vedclass

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(D) Let $M$,$P$,and $C$ be the sets of students who opted for Mathematics,Physics,and Chemistry,respectively.$n(M) = \lfloor \frac{140}{2} \rfloor = 7

Vedclass · Accepted Answer

$38$

Vedclass · Answer

(D) Let $M$,$P$,and $C$ be the sets of students who opted for Mathematics,Physics,and Chemistry,respectively.$n(M) = \lfloor \frac{140}{2} floor = 70$$n(P) = \lfloor \frac{140}{3} floor = 46$$n(C) = \lfloor \frac{140}{5} floor = 28$Now,find the intersections:$n(M \cap P) = \lfloor \frac{140}{	ext{lcm}(2,3)} floor = \lfloor \frac{140}{6} floor = 23$$n(M \cap C) = \lfloor \frac{140}{	ext{lcm}(2,5)} floor = \lfloor \frac{140}{10} floor = 14$$n(P \cap C) = \lfloor \frac{140}{	ext{lcm}(3,5)} floor = \lfloor \frac{140}{15} floor = 9$$n(M \cap P \cap C) = \lfloor \frac{140}{	ext{lcm}(2,3,5)} floor = \lfloor \frac{140}{30} floor = 4$Using the Principle of Inclusion-Exclusion:$n(M \cup P \cup C) = n(M) + n(P) + n(C) - (n(M \cap P) + n(M \cap C) + n(P \cap C)) + n(M \cap P \cap C)$$n(M \cup P \cup C) = 70 + 46 + 28 - (23 + 14 + 9) + 4 = 144 - 46 + 4 = 102$The number of students who did not opt for any course is $140 - 102 = 38$.

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