Let $$(1 + x + x^2)^{10} = a_0 + a_1 x + a_2 x^2 + \ldots + a_{20} x^{20}$$. If $$(a_1 + a_3 + a_5 + \ldots + a_{19}) - 11a_2 = 121k$$, then k is equal to ________.

Let $$(1 + x + x^2)^{10} = \displaystyle\sum_{r=0}^{20} a_r x^r$$.

Substituting $$x = 1$$: $$(1 + 1 + 1)^{10} = 3^{10} = \displaystyle\sum_{r=0}^{20} a_r$$.

Substituting $$x = -1$$: $$(1 - 1 + 1)^{10} = 1 = \displaystyle\sum_{r=0}^{20} a_r(-1)^r$$.

Subtracting: $$3^{10} - 1 = 2(a_1 + a_3 + a_5 + \ldots + a_{19})$$.

So $$a_1 + a_3 + a_5 + \ldots + a_{19} = \dfrac{3^{10} - 1}{2} = \dfrac{59049 - 1}{2} = 29524$$.

Now we need $$a_2$$. The coefficient of $$x^2$$ in $$(1 + x + x^2)^{10}$$:

Using the multinomial expansion $$(1 + x + x^2)^{10} = \displaystyle\sum \dfrac{10!}{i!j!k!}$$ where $$i + j + k = 10$$ and $$j + 2k = 2$$.

Possible values: $$(j, k) = (0, 1)$$ giving $$i = 9$$, or $$(j, k) = (2, 0)$$ giving $$i = 8$$.

$$a_2 = \dfrac{10!}{9!0!1!} + \dfrac{10!}{8!2!0!} = 10 + 45 = 55$$

Now, $$(a_1 + a_3 + \ldots + a_{19}) - 11a_2 = 29524 - 11(55) = 29524 - 605 = 28919$$.

$$28919 = 121k$$, so $$k = \dfrac{28919}{121} = 239$$.

Hence, the correct answer is 239.