To compare the energies of electrons in a multielectron system, we must consider both the principal quantum number ($$\mathrm{n}$$) and the azimuthal quantum number ($$\mathrm{l}$$). In multielectron atoms, the energy levels are influenced by electron-electron repulsions, which modify the energy ordering compared to the hydrogen atom.

The general rule, called the "n+l" rule or Madelung rule, states that the energy of an electron in a multielectron atom is primarily determined by the sum of the principal quantum number ($$\mathrm{n}$$) and the azimuthal quantum number ($$\mathrm{l}$$). An orbital with a lower $$(\mathrm{n} + \mathrm{l})$$ value has lower energy. If two orbitals have the same $$(\mathrm{n} + \mathrm{l})$$ value, the orbital with the lower $$\mathrm{n}$$ value has lower energy.

Now, let's analyze each set of quantum numbers:

(A) $$\mathrm{n}=4, \mathrm{l}=1$$

$$\mathrm{n} + \mathrm{l} = 4 + 1 = 5$$

(B) $$\mathrm{n}=4, \mathrm{l}=2$$

$$\mathrm{n} + \mathrm{l} = 4 + 2 = 6$$

(C) $$\mathrm{n}=3, \mathrm{l}=1$$

$$\mathrm{n} + \mathrm{l} = 3 + 1 = 4$$

(D) $$\mathrm{n}=3, \mathrm{l}=2$$

$$\mathrm{n} + \mathrm{l} = 3 + 2 = 5$$

(E) $$\mathrm{n}=4, \mathrm{l}=0$$

$$\mathrm{n} + \mathrm{l} = 4 + 0 = 4$$

Based on the $$\mathrm{n} + \mathrm{l}$$ values, the order of increasing energy is:

(C) = (E) < (A) = (D) < (B)

Among (C) and (E), (C) has lower $$\mathrm{n}$$ value, so it is lower in energy:

(C) < (E) < (A) = (D) < (B)

Among (A) and (D), (D) has lower $$\mathrm{n}$$ value, so it is lower in energy:

Thus, the correct order is: (C) < (E) < (D) < (A) < (B)

The correct answer is Option C:

$$ (\mathrm{C}) < (\mathrm{E}) < (\mathrm{D}) < (\mathrm{A}) < (\mathrm{B}) $$