Help understand probability in a simply random sample

#1
Quoted is an extract for Sample Survey Principles and Methods, Vic Barnett(2002) Pg 34

The concept of probability averaging only arises in relation to some prescribed probability sampling schemes. Thus, for simple random sampling we have the concept of the expected value of \(y_i\), the \(ith\) observation in the sample. That is,

\(E[y_i] = \sum_{j=1}^{N} Y_j Pr(y_i=Y_j)=\frac{1}{N}\sum_{j=1}^{N} Y_j=\bar{Y}\)

The result that \(Pr(y_i=Y_j) = \frac{1}{N}\) holds because the number of samples with \(y_i=Y_j\) is \(\frac{(N-1)!}{(N-n)!}\) and each has probability \(\frac{(N-n)!}{N!}\)


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I unsure as to justify to myself that \(y_i=Y_j\) is \(\frac{(N-1)!}{(N-n)!}\), the book doesn't discuss this further.

In terms of notation, we have a population \(Y_1, Y_2, ...,Y_N\) and a sample \(y_1, y_2, ...,y_n\)