In a class of 140 students numbered 1 to140 , all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics…

From the given $140$140 students, all even numbered students opted Mathematics $(M),$(M), hence, the number of students opted Mathematics is $M\equiv [\frac{140}{2}]=70,$M≡[2140​]=70, where $[x]$[x] denotes the greatest integer function of $x.$x.

Similarly, the students who opted Physics are $P\equiv [\frac{140}{3}]=46$P≡[3140​]=46

And, the students who opted Chemistry are $C\equiv [\frac{140}{5}]=28$C≡[5140​]=28

Similarly, the students who opted Mathematics and Physics are the numbers which are divisible by both 2&3 i.e. $6,$6, and their number is $[\frac{140}{6}]=23$[6140​]=23

The students who opted Mathematics and Chemistry are the numbers which are divisible by both 2&5 i.e. $10,$10, and their number is $[\frac{140}{10}]=14$[10140​]=14

The students who opted Physics and Chemistry are the numbers which are divisible by both 3&5 i.e. $15,$15, and their number is $[\frac{140}{15}]=9$[15140​]=9

And, the students who opted all the three subjects are the numbers which are divisible by 2,3&5 i.e. $30,$30, and their number is $[\frac{140}{30}]=4$[30140​]=4

Now, using

$$n(P\cup C\cup M)=n(P)+n(C)+n(M)-n(P\cap C)-n(C\cap M)-n(M\cap P)+n(P\cap M\cap C)$$n(P∪C∪M)=n(P)+n(C)+n(M)−n(P∩C)−n(C∩M)−n(M∩P)+n(P∩M∩C)

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

$$=102$$=102

Thus, $102$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 $=140-102=38.$=140−102=38.