Question

Let A and B be sets containing 2 and 4 elements respecetively. The number of subsets $A\times B$ having 3 or more elements is

• A. $219$
• B. $211$
• C. $256$
• D. $220$

Answer 1

The correct option is A $219$

Let $A=\{x,y\}$

$B=\{a,bc,d\}$

$A\times B$ has $2\times 4=8$ elements

Total substance of $A\times B=2^8=256$

$\therefore$Total number of subsets of $A\times B$ having $3$ or more elements

$=256-(1nullset+8singletonset{-}^8{C}_{2}having2elements)$

$=256-1-8-\frac{8!}{6!2!}$

$=256-1-8-\frac{8\times 7\times 6!}{6!2!}$

$=256-1-8-28=219$