# Inclusion Probability in Simple Random Sampling (SRS) Without Replacement

Imagine that we want to choose a sample of size $n$ from a population of size $N$. Let $j$ be a unit contained within the population of size $N$.

What is the inclusion probability of unit $j$ for simple random sampling without replacement? In other words, what is the probability that our sample of size $n$ will contain $j$?

According to Thompson M.E. (1997):

"Since there are ${N-1}\choose{n-1}$ samples of size $n$ which contain $j$ and each has probability $1/{{N}\choose{n}}$, then $\pi_j = {{N-1}\choose{n-1}} \big/{{N}\choose{n}} = \frac{n}{N}$."

I understand that the probability of obtaining any given sample is $1/{{N}\choose{n}}$. However, I don't understand the other claim, which states that there are ${N-1}\choose{n-1}$ samples of size $n$ that contain unit $j$... Any help or guidance would be appreciated. Thanks!

Side note: there is a far easier way to compute $\pi_j$. A simple random sample can be obtained by randomly permuting the $N$ units and choosing the first $n$. It should be clear that the probability that unit $j$ ends up in the first $n$ after permuting is $n/N$.
The number of samples of size $n$ which contain $j$ is simply counting the number of ways you can choose the remaining $n-1$ elements of your sample. There are $N-1$ other units to choose from, so there are $\binom{N-1}{n-1}$ ways.