The number of subshells associated with $$n = 4$$ and $$m = -2$$ quantum numbers is:

We begin by recalling the ranges allowed by the quantum-number rules of the hydrogen-like atom.

First, the azimuthal quantum number $$l$$ can take all integral values from $$0$$ up to $$n-1$$. In symbols, we write the rule

$$0 \le l \le n-1.$$

Second, the magnetic quantum number $$m_l$$ (here simply written as $$m$$) can take every integral value from $$-l$$ to $$+l$$. That rule is written as

$$-l \le m \le +l.$$

Now we substitute the principal quantum number given in the question, namely $$n = 4$$. Using the first rule we obtain every permissible $$l$$ value for this $$n$$:

$$l = 0, \; 1, \; 2, \; 3.$$ (We have four possibilities because $$l$$ must be less than $$n = 4$$.)

The question also specifies the magnetic quantum number $$m = -2$$. From the second rule we know that a particular $$l$$ value is acceptable only if it satisfies

$$|m| \le l.$$

Here $$|m| = |-2| = 2$$, so we must have $$l \ge 2$$. Looking back at the list $$l = 0, 1, 2, 3$$, the values that satisfy $$l \ge 2$$ are

$$l = 2 \quad\text{and}\quad l = 3.$$

Each distinct pair $$(n, l)$$ corresponds to one subshell (for example, $$l = 2$$ with $$n = 4$$ is the $$4d$$ subshell, and $$l = 3$$ with $$n = 4$$ is the $$4f$$ subshell). Because we have exactly two qualifying $$l$$ values, we have two distinct subshells that can possess the magnetic quantum number $$m = -2$$ while still belonging to principal shell $$n = 4$$.

Hence the total number of subshells associated with $$n = 4$$ and $$m = -2$$ is

$$2.$$

Hence, the correct answer is Option B.