Doubt about sampling without replacement

Question

A problem consists of five numbers $2,3,6,8 and 11$. Consider all possible sample space of size 2 that can be drawn with and without replacement from this population then find (a) mean of the population (b) the std deviation of the population (c) the mean of the sampling distributon of means

I was dealing with this problem.. Here for possible sample with replacement of size 2, they have considered both {4,3} and {3,4}.. My question is, aren't they both same sample set?

If I believe in first set we got '4' first and then '3' nd. And in second set they got '3' first and then '4'... in that sense both sets has to be different so different sample space.But in that case, for sampling without replacement also they should have considered {4,3} and {3,4} both...like for first sample they got '4' first and then '3' and for the second sample they got '3' first and then '4'.

Please clear my this doubt...

May be I am wrong with my definitions of with and without replacement... for me with replacement means after selecting one element for the sample you have put it again in the population set to choose the other, so there is chance of getting the same element again...so repetition is allowed in samples...But this is not the case with without replacement

thanks in advance...

I dont know why I am not getting answer, may be I used some pic instead of writing... so I written the question myself and reposting

Answer 1

With replacement, we can draw not only $(3,4)$ and $(4,3)$, but also $(3,3)$ and $(4,4)$. Since each of these four events has the same probability, drawing the set $\{3,4\}$ has twice the probability as the set $\{3,3\}$ (which, as a set, has only one element, but we are interested in the order independence). Since it is easier to handle the four equally probable events rather than three events with different probabilities, we count $(3,4)$ and $(4,3)$ separately.

Without replacement, we only draw $(a,b)$ for $a\ne b$; we don't have $(a,a)$. Thus, we can count $(a,b)$ and $(b,a)$ as $\{a,b\}$, without having to worry about $(a,a)$ having half the probability; we just divide the sample size in half.