Let Number of Students Playing Chess be : n(Ch) = 22

Let Number of Students Playing Carrom be : n(Ca) = 21

Let Number of Students Playing Table Tennis be : n(TT) = 15

Given Number of Students Playing both Chess and Table Tennis = 10

⇒ n(Ch ∩ TT) = 10

Given Number of Students Playing both Carrom and Table Tennis = 8

⇒ n(Ca ∩ TT) = 8

Given Number of Students Playing all the three Games = 6

⇒ n(Ch ∩ Ca ∩ TT) = 6

Given that Each Student in the Class of 35 Plays atleast one Game among Chess, Carrom and Table Tennis

⇒ Total number of Students Playing atleast one Game = 35

⇒ n(Ch ∪ Ca ∪ TT) = 35

We know that :

n(Ch ∪ Ca ∪ TT) = n(Ch) + n(TT) + n(Ca) - n(Ch ∩ TT) - n(Ca ∩ TT) - n(Ch ∩ Ca) + n(Ch ∩ Ca ∩ TT)

Substituting all the Respective values we get :

⇒ 35 = 22 + 15 + 21 - 10 - 8 - n(Ch ∩ Ca) + 6

⇒ 35 = 46 - n(Ch ∩ Ca)

⇒ n(Ch ∩ Ca) = 11

⇒ Number of Students Playing both Carrom and Chess = 11

From the Venn Diagram :

We can Notice that Number of Students Playing both Carrom and Chess also includes the students who are playing all the Three Games.

⇒ Number of Students who are playing both Carrom and Chess but not Table Tennis can be found by Subtracting Number of students playing all the Three games from Number of students playing both Carrom and Chess.

(i) Number of Students Playing both Carrom and Chess but not Table Tennis = n(Ch ∩ Ca) - n(Ch ∩ Ca ∩ TT) = 11 - 6 = 5

(ii) Number of Students playing only Chess will be :n(Ch) - { n(Ch ∩ Ca) - n(Ch ∩ Ca ∩ TT) + n(Ch ∩ Ca ∩ TT) + n(Ch ∩ TT) - n(Ch ∩ Ca ∩ TT) }⇒ 22 - {11 - 6 + 6 + 10 - 6}⇒ 22 - 15 = 7Number of Students only playing Chess will be : 7

(iii) Number of Students playing only Carrom will be :n(Ca) - { n(Ch ∩ Ca) - n(Ch ∩ Ca ∩ TT) + n(Ch ∩ Ca ∩ TT) + n(Ca ∩ TT) - n(Ch ∩ Ca ∩ TT) }⇒ 21 - {11 - 6 + 6 + 8 - 6}⇒ 21 - 13 = 8Number of Students only playing Carrom will be : 8