Given that

$n( X \cup Y )=50, n( X )=28, n( Y )=32$

$n( X \cap Y )=?$

By using the formula

$n( X \cup Y )=n( X )+n( Y )-n( X \cap Y ),$

we find that

$ n( X \cap Y ) =n( X )+n( Y )-n( X \cup Y ) $

$=28+32-50=10 $

Alternatively, suppose $n( X \cap Y )=k,$ then

$n( X – Y )=28-k, n( Y – X )=32-k$ (by Venn diagram in Fig  )

This gives $50=n( X \cup Y )=n( X – Y )+n( X \cap Y )+n( Y – X )$

$=(28-k)+k+(32-k)$

Hence $k=10$