that X ∪ Y has elements, X has elements and Y has elements, how many elements does X ∩ Y have ? Fig . MATHEMATICS Solution Given that n ( X ∪ Y ) = , n ( X ) = , n ( Y ) = , 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 ) = + – = Alternatively , suppose n ( X ∩ Y ) = k , then n ( X – Y ) = – k , n ( Y – X ) = – k (by Venn diagram in Fig .
) This gives = n ( X ∪ Y ) = n (X – Y) + n (X ∩ Y) + n ( Y – X) = ( – k ) + k + ( – k ) Hence k = . Example In a school there are teachers who teach mathematics or physics. Of these, teach mathematics and 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 ) = , n ( M ) = and n ( M ∩ P ) = We wish to determine n ( P ). Using the result n ( M ∪ P ) = n ( M ) + n ( P ) – n ( M ∩ P ), we obtain = + n ( P