The number of ways to distribute 30 identical candies among four children $$C_1, C_2, C_3$$ and $$C_4$$ so that $$C_2$$ receives atleast 4 and atmost 7 candies, $$C_3$$ receives atleast 2 and atmost 6 candies, is equal to

We need to distribute 30 identical candies among four children $$C_1, C_2, C_3, C_4$$ with constraints:

• $$C_1 \geq 0$$, $$C_4 \geq 0$$
• $$4 \leq C_2 \leq 7$$
• $$2 \leq C_3 \leq 6$$

Substitute to remove lower bounds:

Let $$C_2 = 4 + a$$ where $$0 \leq a \leq 3$$, and $$C_3 = 2 + b$$ where $$0 \leq b \leq 4$$.

Then: $$C_1 + C_4 = 30 - (4 + a) - (2 + b) = 24 - a - b$$

Count the number of non-negative integer solutions for $$C_1 + C_4 = 24 - a - b$$:

The number of solutions is $$24 - a - b + 1 = 25 - a - b$$.

Sum over all valid values of $$a$$ and $$b$$:

$$\text{Total} = \sum_{a=0}^{3} \sum_{b=0}^{4} (25 - a - b)$$

For $$a = 0$$: $$(25 + 24 + 23 + 22 + 21) = 115$$

For $$a = 1$$: $$(24 + 23 + 22 + 21 + 20) = 110$$

For $$a = 2$$: $$(23 + 22 + 21 + 20 + 19) = 105$$

For $$a = 3$$: $$(22 + 21 + 20 + 19 + 18) = 100$$

$$\text{Total} = 115 + 110 + 105 + 100 = 430$$

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