I am sharing a proof encountered in William Cochran(1977: page 22): Sampling Techniques.The sample mean $\bar y$ is an unbiased estimate of the population mean $μ$ in a simple random sampling (SRS) process while sampling without replacement. He starts with
$$E(\bar y) = \frac{\sum_{}\bar y}{_N C _n} = \frac{\sum_{} y_1+y_2+y_3+....+y_n}{n[N!/n!(N-n)!]}$$
My explanation of this part is: He calculates expectation where $\dfrac{1}{\binom{N}{n}} $ is probability of choosing a sample.
To evaluate this sum, we find out in how many samples any specific value $y_i$ appears. Since there are $(N-1)$ other units available for the rest of the sample and $(n-1)$ other places left in our sample, the number of samples containing $y_i$ is
This above statements I copied from book and I didn't understand.
$$\overbrace {_{N-1} C _{n-1} = \frac{(N-1)!}{(n-1)!(N-n)!]}}$$
$$\underbrace{\sum_{} y_1+y_2+y_3+....+y_n = \frac{(N-1)!}{(n-1)!(N-n)!]}(y_1+y_2+y_3+....+y_n)}$$
Plug this into the first equation and obtain:
$${E(\bar y) = \frac{(N-1)!}{(n-1)!(N-n)!]} \frac{n!(N-n)!}{nN!}(y_1+y_2+y_3+....+y_n)}$$
This simplifies to:
$$E(\bar y) = \frac{(y_1+y_2+y_3+....+y_n)}{N} $$
This is equal to $\mu$. I don't understand the portion in braces please can someone help me understand it using an example thoroughly.
Here is a screenshot of the proof.
