Number of samples in SRSWR and SRSWOR In a population of N units, if a sample of n units is chosen, the number of possible samples is

*

*N ^ n (SRSWR)


*NCn. (SRSWOR)
Taking an example, let there be a population (a,b,c) and let a sample of 2 units be chosen.
In SRSWOR, there would be 3C2, i.e 3 samples, viz (a,b), (a,c) , (b,c) .
In SRSWR, there would be 3^2 , i.e 9 samples, viz (a,a), (b,b), (c,c), (a,b), (b,a), (a,c), (c,a), (b,c), (c,b).
So why exactly are the samples {(a,b) &(b,a)} , {(b,c) & (c,b)} and {(a,c) & (c,a)} considered distinct in SRSWR but non-distinct in SRSWOR?
PS: If the samples with the same elements but different order are to be considered distinct, shouldnt the number of samples in SRSWOR be NPn (Permutation)? However, most books give the NCn formula.
Kindly help. Thanks.
 A: A reason (not an excuse) for this confusion may be as follows: In sampling with replacement, some people see an implied order of selection. By contrast, in sampling without replacement, one may consider the several choices to be made simultaneously.
When computing probabilities via combinatorial arguments, it is important EITHER to use ordered sampling in both numerator and denominator, OR to use unordered sampling in both.
Esample: An urn has 3 red chips and 2 green ones. You sample two without repl. What is the probability you get one of each color?

*

*Unordered: $${2\choose 1}{3\choose 1}/{5\choose 2} = 6/10 =3/5.$$


*Ordered: $$P(G_1R_2)+P(R_1G_2)= 6/20 + 6/20\\ = 12/20 = 3/5.$$


*By simulation: Computer chooses chips sequentially, but we look only at the
total score, which does not reveal order.
Procedure sample samples without replacement, unless argument repl=T is used.
(With a million draws of two balls, we should
get 2 or 3 place accuracy for the probability.
.
set.seed(513)
urn = c(1,1,1, 2,2,2)  # reds count 1; greens count 2
t = replicate(10^6, sum(sample(urn, 2)))
mean(t==3) # total score 3 if one of each color
[1] 0.599759  # aprx 3/5

