For integers $$n$$ and $$r$$, let $$\binom{n}{r} = \begin{cases} {}^nC_r, & \text{if } n \geq r \geq 0 \\ 0, & \text{otherwise} \end{cases}$$. The maximum value of $$k$$ for which the sum $$\sum_{i=0}^{k} \binom{10}{i}\binom{15}{k-i} + \sum_{i=0}^{k+1} \binom{12}{i}\binom{13}{k+1-i}$$ is maximum, is equal to ______.

Using the Vandermonde identity, the first sum $$\sum_{i=0}^{k} \binom{10}{i}\binom{15}{k-i} = \binom{25}{k}$$, since $$\binom{n}{r} = 0$$ when $$r > n$$ or $$r < 0$$. Similarly, the second sum $$\sum_{i=0}^{k+1} \binom{12}{i}\binom{13}{k+1-i} = \binom{25}{k+1}$$.

So we need to find the maximum value of $$k$$ for which $$\binom{25}{k} + \binom{25}{k+1}$$ is maximized.

By Pascal's identity, $$\binom{25}{k} + \binom{25}{k+1} = \binom{26}{k+1}$$. The binomial coefficient $$\binom{26}{m}$$ is maximized when $$m = 13$$ (since 26 is even, the maximum occurs uniquely at $$m = 13$$). Setting $$k + 1 = 13$$ gives $$k = 12$$.

Therefore, the maximum value of $$k$$ is $$12$$.

The correct answer is $$12$$.