If $S$ and $T$ are two sets such that $S$ has $21$ elements, $T$ has $32$ elements, and $S$ $\cap \,T$ has $11$ elements, how many elements does $S\, \cup$ $T$ have?
It is given that:
$n(S)=21, n(T)=32, n(S \cap T)=11$
We know that:
$n(S \cup T)=n(S)+n(T)-n(S \cap T)$
$\therefore n(S \cup T)=21+32-11=42$
Thus, the set $(S \cup T)$ has $42$ elements.
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
$A-D$
If $X=\{a, b, c, d\}$ and $Y=\{f, b, d, g\},$ find
$X \cap Y$
Let $V =\{a, e, i, o, u\}$ and $B =\{a, i, k, u\} .$ Find $V - B$ and $B - V$
If $A=\{1,2,3,4\}, B=\{3,4,5,6\}, C=\{5,6,7,8\}$ and $D=\{7,8,9,10\} ;$ find
$A \cup B \cup D$
If $A = \{x : x$ is a multiple of $4\}$ and $B = \{x : x$ is a multiple of $6\}$ then $A \cap B$ consists of all multiples of