equation on expectation

The sum $$ E(\bar y)=\frac 1n \frac{1}{\binom Nn}\sum_{i=1}^{\binom Nn}\sum_{i=1}^ny_i. $$ appears to be using the indices of the sums in a non-standard way with an intended meaning that cannot be deduced from the literal formula alone, even if you change one of the index variables from $i$ to $j.$

What I think they have in mind is that if the population has $N$ members, there are $\binom Nn$ subsets of the population that could be selected as a sample of size $n.$ The summation $\sum_{i=1}^{\binom Nn}$ is meant to iterate through that list of subsets, visiting each subset exactly once. Moreover, each subset is equally likely to be selected, so it occurs with probability $1/\binom Nn.$

Now within each subset of $n$ members of the population, we can arbitrarily label the members of the subset as "member number $1$," "member number $2$," up to "member number $n$," and then we can write the observed $y$-values of those members as $y_1, y_2,\ldots y_n$ respectively. Then the sample mean for that particular subset is $\frac 1n \sum_{i=1}^ny_i.$

Notice that this means $y_1$ in one of the $\binom Nn$ samples is not necessarily the same as $y_1$ in one of the other samples. In fact, if $N > n$ and each member of the population has a different $y$-value, it is impossible for $y_1$ to have the same value in each subset.

So I'm confident we are absolutely not meant to take the formula for this sum literally.

A more accurate mathematical notation might be to number each of the members of the population from $1$ to $N$ and assign $y_i$ to the $y$-value of member number $i.$ Number each subset of $n$ members with a number from $1$ to $\binom Nn$ and let $\kappa_j$ be the mapping of the set $\{1,\ldots,n\}$ to the set of indices that tells which members of the population belong to subset number $j.$ For example, if the $j$th subset contains members numbered $k_1,$ $k_2,\ldots, k_n$ with $y$-values $y_{k_1},$ $y_{k_2},\ldots, y_{k_n}$ respectively, then $\kappa_j(i) = k_i.$

Then the first sum can be written $$ E(\bar y)=\frac 1n \frac{1}{\binom Nn} \sum_{j=1}^{\binom Nn}\sum_{i=1}^n y_{\kappa_j(i)}. \tag1 $$

But now let's consider the original "member numbers" of the members of the population, according to which there is a $y$-value $y_m$ for each integer $m$ such that $1\leq m\leq N.$ How many times does that particular subscript of $y$ occur in the summation in Equation $(1),$ summed over all possible values of $j$ and all possible values of $i$?

Given any particular index $m$ whose occurrences we want to count, it should be easy to see that for any $j$ there can be at most one $i$ such that $\kappa_j(i) = m.$ Moreover, there will be one such $i$ only if $m$ is the member number of one of the members of the population that belongs to subset number $j.$ That is, $y_m$ will occur in the sum precisely once for each $n$-member subset of the population that includes member number $m.$ The number of occurrences of $y_m$ is the number of $n$-member subsets that contain member $m$ of the population.

How many $n$-member subsets contain the $m$th member of the population? To form such a subset, we must choose exactly $n-1$ of the remaining $N-1$ members of the population. We can do this in $\binom{N-1}{n-1}$ ways. So that's how many subsets there are. Summing $y_m$ across all its occurences, it contributes $\binom{N-1}{n-1} y_m$ to the total sum.

If we consider this for each $m$ from $1$ to $N,$ we find that the double summation in Equation $(1)$ can be rewritten: $$ \sum_{j=1}^{\binom Nn}\sum_{i=1}^n y_{\kappa_j(i)} = \sum_{m=1}^N \binom{N-1}{n-1} y_m. $$

Since $i$ does not occur on the right-hand side, we can change the index variable there from $m$ to $i$: $$ \sum_{j=1}^{\binom Nn}\sum_{i=1}^n y_{\kappa_j(i)} = \sum_{i=1}^N \binom{N-1}{n-1} y_i. $$

And there you have the desired result.