Am trying to show that $\lim\_{n \rightarrow \infty} \frac{1}{2^{2n}} \sum\_{k=1}^n \sum\_{i=0}^{k-1} \binom{n}{k} \binom{n}{i} =0.5.$ 


I think that the above result is true but am not sure how to prove this. Any help on this will be greatly appreciated. Thanks in advance! 



A combinatorics question: $\lim\limits_{n \to \infty} \frac1{2^{2n}} \sum\limits_{k=1}^n \sum\limits_{i=0}^{k-1} \binom nk \binom ni = \frac12$
We can rewrite the sum as
\begin{align\*}
\sum\_{k=1}^n \sum\_{i=0}^{k-1} \binom{n}{k} \binom{n}{i}
&=\sum\_{0\leq i<k\leq n}\binom{n}{k} \binom{n}{i}\\
&=\frac{1}{2}\sum\_{\substack{0\leq i,k\leq n\\i\neq k}}\binom{n}{k}\binom{n}{i}\\
&=\frac{1}{2}\left(\sum\_{0\leq i\leq n}\binom{n}{i}\right)^2-
\frac{1}{2}\sum\_{0\leq i\leq n}\binom{n}{i}^2\\
&=2^{2n-1}-\frac{1}{2}\binom{2n}{n}.
\end{align\*}
It is well-known that
$$\binom{2n}{n}\sim\frac{2^{2n}}{\sqrt{\pi n}},$$
hence
$$\sum\_{k=1}^n \sum\_{i=0}^{k-1} \binom{n}{k} \binom{n}{i}=\left(\frac{1}{2}-o(1)\right)2^{2n}.$$
The proof is complete.


