Let us take some examples:

Ex1: If X and Y are two sets such that X ∪ Y has 50 elements, X has 28 elements and Y has 32 elements, how many elements does X ∩ Y have ? 

Solution:

Given that

n ( X ∪ Y ) = 50, n ( X ) = 28, n ( Y ) = 32,

n (X ∩ Y) = ?

By using the formula

n ( X ∪ Y ) = n ( X ) + n ( Y ) – n ( X ∩ Y ),

we find that

n ( X ∩ Y ) = n ( X ) + n ( Y ) – n ( X ∪ Y )

= 28 + 32 – 50 = 10

Alternatively, suppose n ( X ∩ Y ) = k, then

n ( X – Y ) = 28 – k , n ( Y – X ) = 32 – k

This gives 50 = n ( X ∪ Y ) = n (X – Y) + n (X ∩ Y) + n ( Y – X)

= ( 28 – k ) + k + (32 – k )

Hence k = 10

Ex2.  In a school there are 20 teachers who teach mathematics or physics. Of these, 12 teach mathematics and 4 teach both physics and mathematics. How many teach physics?

Solution:

Let M denote the set of teachers who teach mathematics and P denote the set of

teachers who teach physics. In the statement of the problem, the word or gives us a

clue of union and the word and gives us a clue of intersection. We, therefore, have

n ( M ∪ P ) = 20 , n ( M ) = 12 and n ( M ∩ P ) = 4

We wish to determine n ( P ).

Using the result

n ( M ∪ P ) = n ( M ) + n ( P ) – n ( M ∩ P ),

We obtain

20 = 12 + n ( P ) – 4

Thus n ( P ) = 12

Hence 12 teachers teach physics.

In this way, we apply union and intersection of two sets derivation in problems.