3 Sets

Set $\mathbb D$ is defined as the set of triplets $(x,y,z)$ where $x$ belongs to $\mathbb A$ , $y$ belongs to $\mathbb B$ , and $z$ belongs to $\mathbb C$ , There are 3 elements in $\mathbb A$ , 12 elements in $\mathbb B$ , 13 elements in $\mathbb C$ . There are 3 possibilities for $x$ , 12 possibilities for $y$ , and 13 possibilities for $z$ . So therefore there are:

$3\times 12\times 13 = \boxed {468} \text{ Elements in } \mathbb D$

say, $$ A = {a₁, a₂, a₃} $$ B = {b₁, b₂, b₃, ....... b₁₂} $$ C = {c₁, c₂, c₃, ....... c₁₃} $$ => (x, y, z) = (a₁, b₁, c₁), (a₁, b₁, c₂), (a₁, b₁, c₂) ............ (a₃, b₁₂, c₁₃) $$ => total number of elements are = 3x12x13 = 468

cardinality of cartesian product of set A and setB is 36 And that of set C and (A,B)is 13*36=468