Difference between permutations and combinations Ok, I know this question has been asked a thousand times. I do understand that permutations mean that order matters where combinations not. I know there's more to it, like what we are counting and the way we are counting, etc. But for some reason I found the wording of the questions really confusing, look for example these two problems extracted from a stats book and please explain why permutations is used in one and combination in the other, why orders matters in one and not the other since they look the same. To facilitate things, instead of 20 and 3 I used 4 and 2 like pick 2 from A, B, C, D for both examples. Why allow AB and BA in example 24 and not in 25?
EXAMPLE 24
An assembler of electronic equipment has 20 integrated-circuit chips on her table, and she must solder three of them as part of a larger component. In how many ways can she choose the three chips for assembly?
Solution
Using Theorem 6, we obtain the result
20P3 = 20!/17! = 20·19·18 = 6,840
EXAMPLE 25
A lot of manufactured goods, presented for sampling inspection, contains 16 units. In how many ways can 4 of the 16 units be selected for inspection?
Solution
According to Theorem 7,
16C4 = 16!/4!12! = 16·15·14·13/4·3·2·1 = 1,092 ways 
 A: It is always important to keep the distinction between ordered and unordered
outcomes in mind. But sometimes you have a choice which kind of outcomes to use.
Sometimes probability problems can be solved in either of two ways, using
permutations (ordered outcomes) throughout, or combinations (unordered outcomes)
throughout.
Here is an example: An urn contains seven balls, 4 red and 3 green. Two balls
are withdrawn without replacement. What is the probability they are both red.
Permutations. (It might help to pretend there are serial numbers on the balls so we can keep track of otherwise indistinguishable red balls.) There are $P(7, 2) = 7(6) = 42$ possible ordered outcomes.
Of these 42, there are $P(4, 2) = 4(3) = 12$ outcomes with two red balls. So
$P(\text{Both R}) = 12/42 = 6/21.$ 
If the balls are numbered from 1 through 7 and red balls have numbers 1 through 4, then we might list the favorable outcomes as follows (where 12 means 1 followed by 2):
12, 13, 14, 23, 24, 34
21, 31,,41, 32, 42, 43 

Combinations. There are $C(7, 2) = {7 \choose 2} = \frac{7!}{2!\cdot 5!} = 21$ possible unordered outcomes. Of these 21, there are $C(4, 2) 
= {4 \choose 2} = 6$ outcomes with two red balls. So
$P(\text{Both R}) = 6/21.$
We might list the favorable outcomes as follows (where 12 means that 1 and 2 were chosen, but without regard to order):
12, 13, 14, 23, 24, 34

Note: Whichever way you choose to work the problem, you must be consistent. If you use permutations to count 'total' (ordered) outcomes in the denominator, then you must choose
permutation to count 'favorable' (ordered) outcomes in the numerator. Similarly, if you consider outcomes as unordered, you must use unordered counts in both numerator and denominator.
