# In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by3 opted Physics course  and those whose number is divisible by 5 opted Chemistry course. Then the number of the students who did not opt for any of the three courses is;Option 1)102Option 2)  42Option 3)  1Option 4)  38

Number of Elements in Union A , B & C -

n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n (A ∩ C) – n (B ∩ C) + n (A ∩ B ∩ C)

- wherein

Given A, B and C be any finite sets. then Number of Elements in union A , B & C is given by this formula.

From the concept,

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

n(B) = student opted physics = 40

n(C) = student opted chemistry = 28

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

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

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

Total no. of student not adopted any course = total - $n\left (A\bigcap B \bigcap C \right )$ = 140 - 102 = 38

Option 1)

102

Option 2)

42

Option 3)

1

Option 4)

38

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