Counting rules - ordered and unordered sample A bag contains n differently coloured balls, If three balls are successively drawn at random, with replacement, what is the probability that at least two balls of the same colour are drawn?
This is a question in my note and the answer is provided. My question is:
the answer said the sample is ORDERED and I dont understand why.
My thought is that if I draw three balls, for example:
red, red, blue
blue, red, red
red, blue, red
They are all the same, because I just need to know what colour and how many of each colour were drawn. In this case, 2 red and 1 blue, and the order does not matter.
Anyone knows why this should be an ordered sample instead of unordered?
 A: Here is a somewhat similar problem (with specific numbers of balls) that illustrates how certain kinds of combinatorial problems might be worked in several ways.

Suppose the bag contains 3 red and 2 green balls, and two balls
  are withdrawn at random without replacement. What is the probability
  that they are of the same color?

Method 1 (General multiplication rule):
$P(R_1R_2) = P(R_1)P(R_2 | R_1) = (3/5)(2/4) = 6/20.$
Similarly, $P(G_1G_2) = 2/20.$ So $P(\text{Same Color})  = 6/20 + 2/20  = 2/5.$
Method 2 (Unordered): $P(\text{Both R}) = {3 \choose 2}/{5 \choose 2} = 3/10.$
Similarly, $P(\text{Both G}) = 1/10,$ so $P(\text{Same Color}) = 3/10 + 1/10 = 2/5.$
Method 3 (Ordered): $P(\text{Both R}) = P(3,2)/P(5,2) = 6/20.$
Similarly, $P(\text{Both G}) = 2/20,$ so 
$P(\text{Same Color}) = 6/20 + 2/20 = 2/5.$
If we choose Method 2, we have a sample space of 10 equally likely unordered outcomes in the denominator and the numerator must also deal with
unordered outcomes. Perhaps red balls are lettered A, B, C and green
balls D and E.  Then the sample space is $$\{AB, AC, AD, AE, BC, BD, BE, CD, CE, DE\},$$ of which AB, AC, BC, and DF are 'favorable'. Notice that in this
unordered sample space AB and BA would be the same outcome; we put letters in
alphabetical order because the real order doesn't matter.
If we chose Method 3, we have twice as many (ordered) outcomes in the sample space
and twice as many 'favorable' ones, so the answer is the same with both methods.
Notice that we have our choice of methods because the event of interest
does not depend on the order of selection. If the question had been "What
is the probability that the second ball drawn is Red?" then Method 2 cannot
be used.
Note: This problem can also be solved using the hypergeometric distribution,
where $X$ is the number of red balls drawn. To answer our question we want
the probability $X$ is either 0 or 2.
In R statistical software, here is how to print the PDF of the particular
hypergeometric distribution required.
x = 0:2;  pdf = dhyper(x, 3, 2, 2) # 3 Red, 2 Non-red, 2 drawn
cbind(x, pdf)
     x pdf
  ## 0 0.1
  ## 1 0.6
  ## 2 0.3

Notes: (1) For sampling with replacement, draws are independent and you could use the binomial distribution.
(2) It is difficult to make sense of unordered samples with replacement for
drawing balls from a bag. (But physicists have discovered that unordered sampling with replacement gives correct answers for some problems in sub-atomic physics.)
