# In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is

Answers (1)

We know,

$\\ n (A \cup B \cup C) = n (A) + n (B) + n (C) - n (A \cap B) - n (A \cap C) - n (B \cap C) + n (A \cap B \cap C)$

According to question.

Let no. of students opted maths = $n(A)$  = 70

student opted physics = $n(B)$ = 40

student opted chemistry = $n(C)$ = 28

$n\left (A\cap B \right ) = 23 , n\left (A\cap C \right ) = 14 , n\left (B\cap C \right ) = 9$

$n\left (A\cap B \cap C \right ) = 4$

$\\ Now , n\left (A\cap B \cap C \right ) = n(A) + n(B) + n(C) - n\left ( A\cap B \right ) - n\left ( A\cap c \right ) - n\left ( B\cap C \right ) + n\left (A\cap B \cap C \right ) = 70 + 40 + 28 - 23 - 14 + 4 = 102$

Therefore, total no. of student not adopting any course = total - $n\left (A\cap B \cap C \right )$ $= 140 - 102 = 38$

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