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

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?

$$N\choose n$$ 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 $$\frac{1}{{N\choose n}n!}=\frac{1}{N (N-1)...(N-(n-1))}$$