In a class of 45 pupils, 21 play chess, 23 play cards and 5 play both the games. Find

(i) how many do not play any of the games;

(ii) how many play chess only;

(iii) how many play cards only.

Total pupils in the class: n(U) = 45

Pupils who play chess: n(C) = 21

Pupils who play cards: n(D) = 23

Pupils who play both games: n(C ∩ D) = 5

(i) how many do not play any of the games

First find n(C ∪ D):

n(C ∪ D) = n(C) + n (D) - n(C ∩ D)

Substituting the values in above, we get:

n(C ∪ D) = 21 + 23 - 5

n(C ∪ D) = 44 - 5

n(C ∪ D) = 39

∴ 39 pupils play both the games.

Pupils who do not play any of the games = n(U) - n(C ∪ D)

Substituting the values in above, we get:

Pupils who do not play any of the games = 45 - 39 = 6

∴ Number of pupils who do not play any of the games = 6.

(ii) how many play chess only

This represents the set C - D, consisting of pupils who play chess but not cards.

We use the formula:

n(C - D) = n(C) - n(C ∩ D)

Substituting the values in above, we get:

n(C - D) = 21 - 5

n(C - D) = 16

∴ 16 pupils play chess only.

(iii) how many play cards only

This represents the set D - C, consisting of pupils who play cards but not chess.

We use the formula:

n(D - C) = n(D) - n(C ∩ D)

Substituting the values in above, we get:

n(D - C) = 23 - 5

n(D - C) = 18

∴ 18 pupils play cards only.