# Each student in a class of 40 plays at least one indoor game chess, carrom and scrabble. 18 play chess, 20 play scrabble and 27 play carrom. 7 play chess and scrabble, 12 play scrabble and carrom and 4 play chess, carrom and scrabble. Find the number of students who play (i) chess and carrom. (ii) chess, carrom but not scrabble.

2
by Deleted account

## Answers

The Brainliest Answer!
• Brainly User
2016-03-28T06:51:27+05:30
Solution

Let A be the set of students who play chess

B be the set of students who play scrabble

C be the set of students who play carrom

Therefore, We are given n(A ∪ B ∪ C) = 40,

n(A) = 18,         n(B) = 20         n(C) = 27,

n(A ∩ B) = 7,     n(C ∩ B) = 12    n(A ∩ B ∩ C) = 4

We have -

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

Therefore, 40 = 18 + 20 + 27 - 7 - 12 - n(C ∩ A) + 4

40 = 69 – 19 - n(C ∩ A)

40 = 50 - n(C ∩ A) n(C ∩ A) = 50 - 40

n(C ∩ A) = 10

Therefore, Number of students who play chess and carrom are 10.

Also, number of students who play chess, carrom and not scrabble.

= n(C ∩ A) - n(A ∩ B ∩ C)

= 10 – 4

= 6

2016-03-28T07:05:56+05:30
Let,

A --> set of students who play chess.

B --> set of students who play scrabble.

C --> set of students who play carrom

n(A ∪ B ∪ C) = 40,

n(A) = 18,         n(B) = 20         n(C) = 27,

n(A ∩ B) = 7,     n(C ∩ B) = 12    n(A ∩ B ∩ C) = 4

i) We know that,

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)
40 = 18 + 20 + 27 - 7 - 12 - n(C ∩ A) + 4
40 = 50 - n(C ∩ A)
40 = 50 - n(C ∩ A) n(C ∩ A) = 50 - 40
n(C ∩ A) = 10
Number of students who play chess and carrom = 10.
ii) number of students who play chess, carrom but not scrabble.
= n(C ∩ A) - n(A ∩ B ∩ C)
= 10 – 4
= 6