In a class of 140 students numbered 1 to 140 , all even numbered...

From the given $140$ students, all even numbered students opted Mathematics \left(M\right),$hence, the number of students opted Mathematics is$M ≡ \left[\dfrac{140}{2}\right] = 70,$where \left[x\right]$ denotes the greatest integer function of $x .$

Similarly, the students who opted Physics are $P ≡ \left[\dfrac{140}{3}\right] = 46$

And, the students who opted Chemistry are $C ≡ \left[\dfrac{140}{5}\right] = 28$

Similarly, the students who opted Mathematics and Physics are the numbers which are divisible by both $2 \text{ and } 3$ i.e. $6,$ and their number is \left[\dfrac{140}{6}\right] = 23$</p><p>The students who opted Mathematics and Chemistry are the numbers which are divisible by both$2 \text{ and } 5$i.e.$10,$and their number is \left[\dfrac{140}{10}\right] = 14$

The students who opted Physics and Chemistry are the numbers which are divisible by both $3 \text{ and } 5$ i.e. $15,$ and their number is \left[\dfrac{140}{15}\right] = 9$</p><p>And, the students who opted all the three subjects are the numbers which are divisible by$2, 3 \text{ and } 5$i.e.$30,$and their number is \left[\dfrac{140}{30}\right] = 4$

Now, using
$n\left(P \cup C \cup M\right) = n\left(P\right) + n\left(C\right) + n\left(M\right) - n\left(P \cap C\right) - n\left(C \cap M\right) - n\left(M \cap P\right) + n\left(P \cap M \cap C\right)$

$= 46 + 28 + 70 - 9 - 14 - 23 + 4$

$= 102$

Thus, $102$ students have opted for atleast one of the three subjects.

Hence, number of students who did not opt for any of the three courses = 140 - 102 = 38 .