If $X$ and $Y$ are two sets such that $X \cup Y$ has $18$ elements, $X$ has $8$ elements and $Y$ has $15$ elements ; how many elements does $X \cap Y$ have?
It is given that:
$n(X \cup Y)=18, n(X)=8, n(Y)=15$
$n(X \cap Y)=?$
We know that:
$n(X \cup Y)=n(X)+n(Y)-n(X \cap Y)$
$\therefore 18=8+15-n(X \cap Y)$
$\Rightarrow n(X \cap Y)=23-18=5$
$\therefore n(X \cap Y)=5$
If $A=\{3,6,9,12,15,18,21\}, B=\{4,8,12,16,20\},$ $C=\{2,4,6,8,10,12,14,16\}, D=\{5,10,15,20\} ;$ find
$B-D$
If $A=\{3,6,9,12,15,18,21\}, B=\{4,8,12,16,20\},$ $C=\{2,4,6,8,10,12,14,16\}, D=\{5,10,15,20\} ;$ find
$A-C$
Find the union of each of the following pairs of sets :
$A=\{a, e, i, o, u\} B=\{a, b, c\}$
Using that for any sets $\mathrm{A}$ and $\mathrm{B},$
$A \cup(A \cap B)=A$
Let $A, B$ and $C$ be sets such that $\phi \ne A \cap B \subseteq C$. Then which of the following statements is not true ?