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I will illustrate my question with an example. Suppose that there are $14$ balls in a box of which 8 are Blue, $4$ are Yellow, and $2$ are Orange. Suppose we pick two balls from the box and we want to find the probability table of the possible combinations. In this case the combinations (macrostates) possible are: $$OO, OY, OB, YY, YB, BB.$$

Now lets separate the problem into two cases.

$1)$We pick the balls simultaneously.

$2)$We pick the balls one after the other.

In the first case the order of choosing does not matter so, for example, in the macrostate $OY$ it seems that there is only one microstate contributing - $OY$. However in case $2)$ for the same macrostate $OY$ is it correct to say that there are now two microstates contributing $OY,YO$?

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    You can have it either way. If you define $OY$ and $YO$ to be different outcomes, then of course you’re forced to choose sequentially. If you define them to be the same outcome, then you can choose sequentially or simultaneously. – 2017-01-02
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    So the probability depends on my own definition? This doesn't make sense to me. @BrianM.Scott – 2017-01-02
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    The probability depends on what the possible outcomes are. If you define $OY$ and $YO$ to be different outcomes, clearly you’re thinking about making sequential choices or in some other way distinguishing the two draws (e.g., by drawing one with each hand and listing the lefthand draw first). If you define $OY$ and $YO$ to be the same outcome, then you’re clearly not interested in distinguishing order, even if you actually make your two draws sequentially. It’s the space of outcomes that matters; the drawing mechanism is determined by the space of outcomes, not the reverse. – 2017-01-02
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    OK . This makes sense. Thank you – 2017-01-02
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    You’re welcome. – 2017-01-02
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    "Suppose we pick two balls from the box and we want to find the probability of obtaining all the different combinations" - this probability is $0$ **by definition**. How could you possibly get all $6$ combinations by picking only $2$ balls? You need to rephrase your question IMO. – 2017-01-02

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Perhaps it is best to start with a slightly simpler problem. We have an urn containing 5 balls: 3 red and 2 green. We will withdraw two balls at random without replacement. (a) We want the distribution of the random variable $X$ that counts the number of red balls drawn. (b) Also, we want to know the probability that the two balls drawn are of the same color.

Unordered Outcomes. Notice that neither (a) nor (b) mentions the order in which the two balls are drawn. For (a), we have $$P(X = i) = \frac{{3 \choose i}{2 \choose 2-i}}{{5 \choose 2}},$$ for $i = 0,1,2.$ The counts in denominator and numerator are both of unordered outcomes. You may recognize this as a hypergeometric distribution. We can get the table for the PDF of $X$ by simple arithmetic to deal with the factorials involved or by using R statistical software:

i = 0:2;  pdf = dhyper(i, 3, 2, 2)
cbind(i, pdf)  # 'cbind' binds together column vectors
##      i pdf
##      0 0.1
##      1 0.6
##      2 0.3

We can answer (b) as $P(\text{Same Color}) = P(X = 0) + P(X = 2) = 0.4.$

One possible sample space has unordered oucomes $\{GG, RG, RR\},$ with unequal probabilities, each corresponding to one of the three possible values of $X.$

Another possible sample space starts by naming the five balls as r1, r2, r3, g1, g2 and then lists unordered outcomes, each of which shows two balls. Because 'real' order makes no difference, I choose a convenient order of for each of the ${5 \choose 2} = 10$ equally likely outcomes (for example, r1r2, not r2r1).

 r1r2 r1r3 r1g1 r1g2
      r2r3 r2g1 r2g2
           r3g1 r3g2
                g1g2

Ordered Outcomes. If I am going to ask questions for which order makes a difference, then I need to deal with order throughout. For example, what is the probability that a red ball is chosen on the second draw?

One approach is to use conditional probability:

$P(R_1 \cap R_2) = P(R_1)P(R_2 | R_1) = \frac{3}{5}\cdot\frac{2}{4} = 0.3$

$P(G_1 \cap R_2) = P(G_1)P(R_2 | G_1) = \frac{2}{5}\cdot\frac{3}{4} = 0.3$

$P(R_2) = P(R_1 \cap R_2) + P(G_1 \cap R_2) = 0.6.$

Also, by symmetry, we should have $P(R_2) = P(R_1) = 3/5 = 0.6.$

You may want to write out the sample space of 20 equally likely ordered outcomes, which has the ten listed above, plus another ten with the order reversed. (Now, for example, both r1r2 and r2r1 need to be listed.)

Questions in which order is not important can be answered using ordered outcomes, but not the reverse. For example, we could find $$P(\text{Same Color}) = P(R_1 \cap R_2) + P(G_1 \cap G_2) = 0.3 + 0.1 = 0.4.$$

The most important lesson here is to decide at the start whether you are going to use ordered or unordered outcomes, and then stay with that decision throughout the computation.