Question

college awarded 38 medals in football 15in basketball 20 in cricket. if these medals went to a total of 58 men only 3men got medals in all 3 sports how many received medals in exactly two of them​

Answer 1

n(FUBUC) = n(F) + n(B) + n(C) - n(BnC) - n(FnB) - n(FnC) + n(FnBnC)

 

58 = 38 + 15 + 20 - [n(BnC) + n(FnB) + n(FnC)] + 3

 

18 = [n(BnC) + n(FnB) + n(FnC)]

 

Number of men who received medals in exactly two of the three sports

=n(FnBnC') + n(FnB'nC) + n(F'nBnC)

=n(BnC) + n(FnB) + n(FnC) - 3n(FnBnC)

=9   {Since n(FnBnC') = n(FnB) - n(FnBnC)}

Answer 2

Solution :

Let,

V = set of students getting medals in volleyball

F = set of students getting medals in football

B = set of students getting medals in basket ball

We are given the following:

n ( V ) = 38,

n ( F ) = 15

n ( B ) = 20

n ( V U F U B ) = 58

n ( V ∩ F ∩ B ) = 3

Let,

a = Number of students getting medals in volleyball and football but not the basketball

b = Number of students getting medals in football and basketball but not volleyball

c = Number of students getting medals in basketball and volleyball but not football

So,

n ( V U F U B ) is as follows :

⇒ [ 38 - ( a + c + 3 ) ] + [ 15 - ( a + b + 3 ) ] + [ 20 - ( b + c + 3 ) ] + a + b + c = 58

⇒ 67 - ( a + b + c ) = 58

⇒ a + b + c = 9

∴ Number of players who received medals in exactly two of the three sports,

⇒ a + b + c

⇒ 9

∴ Number of players who received medals in exactly two of the three sports is 9 respectively.