<s>[INST] Find $\sum_{i=10}^n (\sum_{r=10}^i \binom{30}{r}\binom{20}{r-10})$=? Evaluate:
$$\sum_{i=10}^n \left(\sum_{r=10}^i \binom{30}{r}\binom{20}{r-10}\right)$$
I tried this and my result come:
$$n\binom{30}{10}\binom{20}{1}+ (n-1)\binom{30}{11}\binom{20}{2} + (n-2)\binom{30}{12}\binom{20}{3} + ...... +(n-(n-1))\binom{30}{n-1}\binom{20}{(n-1)-10} + (n-(n))\binom{30}{n}\binom{20}{n-10}$$
I'm not getting what to do further.
 [/INST] Note 
$$
\sum_{i=10}^n \left(\sum_{r=10}^i \binom{30}{r}\binom{20}{r-10}\right) = \\
\sum_{i=10}^n \left(\sum_{k=0}^{i-10 } \binom{30}{k+10}\binom{20}{k}\right)= \\
\sum_{j=0}^{n-10} \left(\sum_{k=0}^{j} \binom{30}{k+10}\binom{20}{k}\right)= \\
\sum_{j=0}^{20} \left(\sum_{k=0}^{j} \binom{30}{k+10}\binom{20}{k}\right) = \\
\sum_{j=0}^{20} \left(\sum_{k=0}^{j} \binom{30}{20-k}\binom{20}{20-k}\right)
$$
Observe that you can rearrange $\sum_{j=0}^{20}\sum_{k=0}^{j} = \sum_{k=0}^{20} \sum_{j=k}^{20}$ which can be used: 
$$
\cdots = 
\sum_{k=0}^{20} \left(\sum_{j=k}^{20} \binom{30}{20-k}\binom{20}{20-k}\right) = \\
\sum_{k=0}^{20} (20-k) \binom{30}{20-k}\binom{20}{20-k} = \\
30 \sum_{k=0}^{19}  \binom{29}{19-k}\binom{20}{20-k} = \\
 30 \cdot 18851684897584 = 565550546927520
$$
</s>