For what value of $m$ the is sum $\sum_{i = 0}^{m} {10 \choose i}{20 \choose m - i}$ where ${p\choose q} = 0$, if $p<q$, a maximum

Vandermonde's Identity says that $$ \sum_{i=0}^m\binom{10}{i}\binom{20}{m-i}=\binom{30}{m}\tag1 $$ The central binomial coefficient is the greatest. Therefore, the maximum of $(1)$ is when $m=15$; that is, $\binom{30}{15}$.

Here's how I'm gonna to approach it. Hopefully, it'll give some intuition into the Vandermonde identity that you can read up on later.

Consider a jar of $r$ red marbles and $b$ blue marbles. We need to find the number of ways to pick $m$ marbles.

Well we can pick $1$ red and $m - 1$ blue marbles or $2$ red and $m - 2$ blue marbles or $3$ red and $m - 3$ and so forth...

We need to find the sum total of the number of ways to pick $i$ red and $m - i$ blue marbles, which is exactly the RHS of: $$ {r+b \choose m} = \sum_i^m {r \choose i}{b \choose m - i}$$

And the LHS is exactly as we described previously: the number of ways to choose $m$ marbles from all $r + b$ marbles. Does this make sense? We are essentially a combinatorial trick called "counting by two ways".

Now where does ${r+b \choose m}$ achieve its maximum? Well Pascal's triangle has it's maximums of every row down it's center column, so ${r+b \choose m}$ is maximized at $\left\lfloor {\frac{r+b}{2}}\right\rfloor$