The color of lanthanoid ions in solutions is mainly due to the electronic transitions within the 4f subshell. The lanthanoid ions are more likely to be colorless when they have fully filled (with 14 electrons) or completely empty (with 0 electrons) f orbitals because, in these cases, there are no electrons to undergo f-f transitions, and as a result, no absorption of visible light occurs leading to colorlessness.

Let's consider the electronic configurations of the lanthanoid ions provided:

- $$\mathrm{Eu}^{3+}$$: Europium (Eu) has an atomic number of 63. Neutral europium ([Xe]4f⁷ 6s^2) loses three electrons to form Eu³⁺, leaving it with an electronic configuration equivalent to [Xe]4f⁶. With 6 electrons in the f orbital, it can undergo f-f transitions, thus it is not colorless.

- $$\mathrm{Lu}^{3+}$$: Lutetium (Lu) has an atomic number of 71. In its 3+ ionic state, lutetium has lost its 6s and 5d electrons and is left with a completely filled 4f orbital ([Xe] 4f¹⁴). This configuration cannot allow for any f-f transitions, as there are no available energy levels within the f orbital for an electron to jump to, making Lu³⁺ colorless.

- $$\mathrm{Nd}^{3+}$$: Neodymium (Nd) has an atomic number of 60. In its 3+ state ([Xe] 4f³), it clearly has partially filled f orbitals, which can absorb visible light for f-f transitions, so it is not colorless.

- $$\mathrm{La}^{3+}$$: Lanthanum (La) has an atomic number of 57. In its 3+ ionic state, it has a configuration of [Xe], meaning that its 4f orbital is completely empty. Since there are no electrons in the f orbital to undergo f-f transitions, La³⁺ is colorless.

- $$\mathrm{Sm}^{3+}$$: Samarium (Sm) has an atomic number of 62. As a 3+ ion ([Xe] 4f⁵), it too has electrons in the f orbital capable of undergoing f-f transitions, so it is not colorless.

From the analysis, the colorless lanthanoid ions among the ones listed are $$\mathrm{Lu}^{3+}$$ and $$\mathrm{La}^{3+}$$.

Therefore, the number of colorless lanthanoid ions among the given options is 2.