The remainder when $$(2021)^{2022} + (2022)^{2021}$$ is divided by $$7$$ is

We need to distribute 30 identical candies among $$C_1, C_2, C_3, C_4$$ with $$4 \leq C_2 \leq 7$$ and $$2 \leq C_3 \leq 6$$, where $$C_1 \geq 0$$ and $$C_4 \geq 0$$.

For each valid pair $$(C_2, C_3)$$, the remaining $$30 - C_2 - C_3$$ candies are split between $$C_1$$ and $$C_4$$ (both $$\geq 0$$), giving $$30 - C_2 - C_3 + 1 = 31 - C_2 - C_3$$ ways.

We sum over all valid $$(C_2, C_3)$$ pairs:

For $$C_2 = 4$$: $$\displaystyle\sum_{C_3=2}^{6}(27 - C_3) = 25 + 24 + 23 + 22 + 21 = 115$$

For $$C_2 = 5$$: $$\displaystyle\sum_{C_3=2}^{6}(26 - C_3) = 24 + 23 + 22 + 21 + 20 = 110$$

For $$C_2 = 6$$: $$\displaystyle\sum_{C_3=2}^{6}(25 - C_3) = 23 + 22 + 21 + 20 + 19 = 105$$

For $$C_2 = 7$$: $$\displaystyle\sum_{C_3=2}^{6}(24 - C_3) = 22 + 21 + 20 + 19 + 18 = 100$$

Total $$= 115 + 110 + 105 + 100 = 430$$.

The answer is Option D: $$430$$.