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

We begin by noting the basic facts given in the question. The set $$A$$ contains $$2$$ elements, and the set $$B$$ contains $$4$$ elements. Whenever we form the Cartesian product $$A \times B$$, each element of $$A$$ pairs with every element of $$B$$.

By definition of Cartesian product, the total number of ordered pairs in $$A \times B$$ equals the product of the individual cardinalities:

$$|A \times B| \;=\; |A| \times |B|.$$

Substituting the given sizes, we obtain

$$|A \times B| \;=\; 2 \times 4 \;=\; 8.$$

So, the set $$A \times B$$ has exactly $$8$$ elements.

Next, we recall a standard counting fact: a finite set with $$n$$ elements possesses $$2^n$$ distinct subsets. Stating this formally, if a set has $$n$$ elements, then

$$\text{Number of all subsets} \;=\; 2^n.$$

Applying this to our situation with $$n = 8$$, we have

$$\text{Total subsets of }A \times B \;=\; 2^8 \;=\; 256.$$

However, the problem does not ask for all subsets. It asks only for those subsets that contain three or more elements. Therefore we must exclude the subsets of sizes $$0$$, $$1$$, and $$2$$ from the grand total of $$256$$.

To exclude them accurately, we use the combination formula. The number of ways to choose $$k$$ elements from an $$n$$-element set is given by the binomial coefficient

$$\binom{n}{k} \;=\; \frac{n!}{k!\,(n-k)!}.$$

For our set with $$n = 8$$, we compute:

$$ \begin{aligned} \binom{8}{0} &= 1, \\ \binom{8}{1} &= 8, \\ \binom{8}{2} &= 28. \end{aligned} $$

The total number of “small” subsets (those having fewer than $$3$$ elements) is therefore

$$\binom{8}{0} \;+\; \binom{8}{1} \;+\; \binom{8}{2} \;=\; 1 + 8 + 28 \;=\; 37.$$

Now we subtract these $$37$$ unwanted subsets from the complete collection of $$256$$ subsets:

$$256 \;-\; 37 \;=\; 219.$$

This final result, $$219$$, is precisely the number of subsets of $$A \times B$$ that contain $$3$$ or more elements.

Hence, the correct answer is Option A.