If $X$ and $Y$ are two sets such that $X \cup Y$ has $50$ elements, $X$ has $28$ elements and $Y$ has $32$ elements, how many elements does $X$ $\cap$ $Y$ have?
If $X$ and $Y$ are two sets such that $X \cup Y$ has $50$ elements, $X$ has $28$ elements and $Y$ has $32$ elements, how many elements does $X$ $\cap$ $Y$ have?
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 $k=10$
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