Question 1
If A and B are two sets such that A ∪ B has 50 elements, A has 28 elements and B has 42 elements, how many elements does A ∩ B have
Solution
\(n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)\)
50=28+42-x
x=20
Question 2
Sets A and B have 12 and 8 elements respectively. What can be the minimum number of elements in $A \cup B$
Solution
We know that
\(n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)\)
$A \cup B = 12 + 8 - n(A \cap B)$
Now maximum number of elments possible in $A \cap B $ = 8 ( B is subset of A)
So, Minimum number of elements in $A \cup B$ = 12 + 8 -8 =12
Question 3
A college awarded 38 medals in football, 15 in basketball and 20 in cricket. If these medals went to a total of 58 men and only three men got medals in all
the three sports, how many received medals in exactly two of the three sports ?
Solution
Let F, B and C denote the set of men who received medals in football, basketball and cricket, respectively.
Then ,According to the questions, we have
n ( F ) = 38, n ( B ) = 15, n ( C ) = 20 ,$n (F \cup B \cup C ) = 58$ and $ (F \cap B \cap C ) = 3$
Now,
$n (F \cup B \cup C ) = n ( F ) + n ( B )+ n ( C ) - n (F \cap B ) - n (F \cap C ) - n (B \cap C ) \\
+ n ( F \cap B \cap C )$
$58=38+15+20 -(n ( F \cap B ) + n ( F \cap C ) + n ( B \cap C ) ) +3$
or $n ( F \cap B ) + n ( F \cap C ) + n ( B \cap C ) = 18$ -(1)
Now the question is asking for exactly two of the three sports ie.
$n (( F \cap B ) -C) + n (( F \cap C ) -B) + n (( B \cap C )-F)$
=$n ( F \cap B ) + n ( F \cap C ) + n ( B \cap C ) -3 (F \cap B \cap C )$
=18 -9 =9
We can understand this with Venn diagram also
Here, x denotes the number of men who got medals in football and basketball only, y denotes the number of men who got medals in football and cricket only, z denotes the number of men who got medals in basket ball and cricket only and k denotes the number of men who got medal in all the three.
Now ,
$F \cap B = x+k$
$F \cap C = y+k$
$B \cap C = z+k$
We need to find a+b+c
Now as per equation (1)
x +k + y + k + z + k = 18
Therefore a + b + c = 9