Let $$A$$ and $$B$$ be two sets containing four and two elements respectively. Then the number of subsets of the set $$A \times B$$, each having at least three elements is

We have two finite sets, namely $$A$$ and $$B$$, whose cardinalities are given to be $$|A| = 4$$ and $$|B| = 2$$ respectively.

First, recall the definition of the Cartesian product of two sets.

For any sets $$X$$ and $$Y$$, the Cartesian product $$X \times Y$$ is defined as $$\{(x,y) \mid x \in X,\, y \in Y\}$$.

The number of ordered pairs in $$X \times Y$$ is obtained by the fundamental principle of counting: every element of $$X$$ can be paired with every element of $$Y$$, so the total number of pairs is the product $$|X| \cdot |Y|$$.

Applying this to $$A$$ and $$B$$, we obtain

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

Thus there are exactly $$8$$ distinct ordered pairs in $$A \times B$$. Consequently, $$A \times B$$ is a set with eight elements.

Next, we are asked to count the subsets of $$A \times B$$ that have at least three elements.

For a set with $$n$$ elements, the total number of all possible subsets (including the empty set) is given by the power set formula $$2^{n}$$.

Here, since $$n = 8$$, the total number of all subsets of $$A \times B$$ is

$$2^{8} = 256.$$

However, this total counts subsets of every possible size, from $$0$$ up to $$8$$. We do not want every subset; we only want those subsets whose size (cardinality) is at least $$3$$. Hence we must subtract the numbers of subsets that have fewer than $$3$$ elements, namely subsets with $$0$$, $$1$$, or $$2$$ elements.

To do that, we employ the binomial coefficient $$\binom{n}{k}$$, which counts the number of subsets of size $$k$$ from an $$n$$-element set.

We therefore compute:

$$\binom{8}{0}$$ gives the number of empty subsets, which is $$1$$.

$$\binom{8}{1}$$ gives the number of single-element subsets, which is $$8$$.

$$\binom{8}{2}$$ gives the number of two-element subsets. Using the combination formula $$\binom{n}{2} = \dfrac{n(n-1)}{2}$$, we obtain

$$\binom{8}{2} = \dfrac{8 \times 7}{2} = 28.$$

Now we add these three quantities to find the total number of subsets that have fewer than three elements:

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

Finally, we subtract this result from the grand total of $$256$$ subsets to get the desired count of subsets with at least three elements:

$$256 - 37 = 219.$$

Therefore, the number of subsets of $$A \times B$$ each having at least three elements is $$219$$.

Hence, the correct answer is Option B.