In an elementary number theory book I'm reading, there is a comparison of, for r $\ge$ 6
$\binom{3 \cdot 2^{2r}}{3 \cdot 2^{2r-1}}$ and $\binom{2^{2r}}{2^{2r-1}}\binom{2^{2r-1}}{2^{2r-2}}...\binom{2}{1} [ \binom{2^{r+1}}{2^r}\binom{2^r}{2^{r-1}}...\binom{2}{1}]^{2r} $.
I interpret the right side as choosing a half objects from $2^{2r}$, and choosing another half from the half, ..., until I get 1. And simultaneously choosing a half objects from $2^{r+1}$, and choosing another half from this half, ..., until I get 1, for $2r$ times simultaneously.
It is written in the book that the former is bigger than the latter, by considering for r $\ge$ 6
$3 \cdot 2^{2r} \gt (2^{2r}+2^{2r-1}+...+2)+2r(2^{r+1}+2^r+...+2)$.
Since this comment, "observing (the inequality) and interpreting $\binom{2n}{n}$ as the number of ways of choosing n objects from 2n, we can conclude that (the former is indeed bigger than the latter)", is added to this inequality,
I wonder if there is any way of interpreting summation of the upper part numbers in combination notations in the right side with counting the ways of choosing the halves? Also any way of interpreting it with the upper part number in the combination notation in the left side?
And I want to know about any formula which uses such interpretation.
