Probability of selection of a sample in simple random sampling with and without replacement

Question

I am interested in computing the probability of selecting a specific sample say $u_{1}, u_{2},...,u_{n}$ from a population having $N$ sampling units. I know that in without replacement there are total number of possible sample as $N \choose n$ and hence the probability of such selection will be inverse of this.

Similarly, in the with replacement case, the no of such samples will be $N^{n}$ and consequently the probability of such selection will be inverse of this.

But if I try to view this problem as computing the $P(u_{1}, u_{2},...,u_{n})$. I can write:

$P(u_{1}, u_{2},...,u_{n}) = P(u_{1})P(u_{2})....P(u_{n}) $. Let us call it Equation (1)

In the case of with replacement, $P(u_{i}) = \frac{1}{N}$. Putting this in equation (1), It gives us the same result as the previous method gave. But in the situation of Without replacement, we have:

$P(u_{i}) = \frac{1}{N - (i-1)}$.

It is not giving the result which is consistent with the previous one.

Can someone suggest me what is it that I am doing wrong here?

Answer 1

In the first way, you are not considering the order, but in the second way, you are considering the order that they are chosen.
$N\choose n$ is the number of ways to pick a sample of size $n$ from the population without replacement. But, then there are $n!$ different ways to arrange those objects. So, the probability of selecting them in that order is $$\frac{1}{{N\choose n}n!}=\frac{1}{N (N-1)...(N-(n-1))}$$