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 $pFor what value of $m$ the is sum
$$\sum_{i = 0}^{m} {10 \choose i}{20 \choose m - i}\text{where ${p\choose q}$} = 0\text{, if $p<q$, a maximum}$$
My approach
$$\sum_i^{m} {10 \choose i}{20 \choose m - i} = {10 \choose 0}{20 \choose m} + {10 \choose 1}{20 \choose m - 1} + \dots + {10 \choose m}{20 \choose 0}$$
$$(1 +x)^{20} = {20 \choose 0} + {20 \choose 1}x + \dots + {20 \choose m-1}x^{m-1} + {20 \choose m}x^{m} + \dots + {20 \choose 20}x^{20}$$
$$(1 +x)^{10} = {10 \choose 0} + {10 \choose 1}x + \dots + {10 \choose 10}x^{10}$$
Later what to do??
Any other method or hint will be greatly welcomed.

$$(1+x)^{10}= 10_{C_{0}}+10_{C_{1}} x+\cdots+10_{C_{10}} x^{10} \ldots \ldots \ldots \ldots \ldots .(1)$$
$$(1+x)^{20}= 20_{C_{20}}x^{20}+20_{C_{19}} x^{19}+\cdots+20_{C_{0}} \ldots \ldots \ldots \ldots \ldots .(2)$$
On multiplying these two...
$$(1+x)^ {30}= (\cdots)x^{30}+(\cdots)x^{29}+\cdots$$
The largest coefficient would be that of $x^{15}$ and that would be equal to ...
$$\sum_{i=0}^{m}\left(\begin{array}{l}10 \\ i\end{array}\right)\left(\begin{array}{l}20 \\ 15-i\end{array}\right),$$
Hope this helps!
