Question

please answer the question
write full solution​

Answer 1

Answer:

1 is the correct answer I guess

Answer 2

Answer:

Option 3 is the answer

Step-by-step explanation:

Total number of students=60

No.of students who do not play any game =15

No.of students who play games =60-15=45

No.of students who play hockey =n(H)=23

No.of students who play basketball=n(B)=15

No.of students who play cricket=n(C)=20

No.of students who paly both hockey and basketball=n(HnB)=7

No.of students who play both basketball and Cricket=n(BnC)=5

No.of students who play both Hockey and Cricket=n(HnC)=4

Let no.of students who play all the three games be X =n(HnBnC)=X

We know that

n(HUBUC)=n(H)+n(B)+n(C)- n(HnB)- n(BnC)- n(HnC) +n(HnBnC)

45=23+15+20-7-5-4+X

=>45=58-16+X

=>45=42+X

=>45-42=X

=>X=3

No.of students who play all the three games =3

No.of students who play only hockey

=23-4-1-3=15

No.of students who play hockey and cricket but not Basketball = n(HnC)-n(HnBnC)=4-3=1