12 books shelf and bag. I got two varieties for the same question:


*

*Ways that four books out of a bag of 12 books can be placed on a shelf.

*Ways to choose 4 books out of 12 arranged on a shelf and put them in a bag.


Answer for the first one is $12 * 11 * 10 * 9$ but for the second one the answer is $(12 * 11 * 10 * 9) / (4 * 3 * 2 * 1)$ since we don't care about the order.
I am satisfied with the first solution but the solution for the second question opens a door for confusions for me. If I understand correctly in the first question it is assumed that order is important and in the second case it is not. Anyways I will try to think backwards.
My thought process for the first questions is:

There are sequence of choices made after another with the remaining
  choice decreasing in each step (I mentally see it as a tree). So, for
  the first book I have 12 choices, next for the second level of the
  tree I have 11 choices and so on, since we need four choices we would
  go 4 levels deep. The total number of leaves would give us the
  required solution which is $12 * 11 * 10 * 9$.
Lets, investigate more. The leaves represents total number of choices
  and if we see as a tree then each leaf would contain the nodes from
  all the ancestors but in different orders hence there would be
  repetition for eg: (1, 2, 3, 4), (1, 3, 2, 4) and so on, now it
  depends on the question if it needs to consider theses leaves equal or
  different (ordered I think?).

For the second question:

Lets apply the same thought process again. The reasoning seems to work
  fine till first half of my thought process discussed above in the
  first half. But the solution divides the choices by some series of
  multiplication so, basically the author has reduced the number of
  available choices hence it must have removed some duplicates, the only
  duplicates(equivalence classes as told in the source) I see here are
  the repeated tree leaves if I see them as a set hence it means we
  don't need duplication here.

Question:
Am I correct in reasoning for the above solutions and creating the correct mental models?
Ref:
http://ocw.mit.edu/high-school/mathematics/combinatorics-the-fine-art-of-counting/lecture-notes/MITHFH_lecturenotes_2.pdf
PS: This might be unrelated but seems important to me.
In some places people try to solve some problems by reducing the problem to a sequence of bits and choosing all the possible number of places for 1's, how to apply and understand this rule? 
 A: Let's simplify this to choosing $2$ books from $3$ 


*

*Putting them on a shelf, if the books in the bag are $\{A,B,C\}$ then the possibilities are: 


*

*$(A,B)$

*$(A,C)$

*$(B,C)$

*$(B,A)$

*$(C,A)$

*$(C,B)$


*Putting them in a bag, if the books on the shelf are $(A,B,C)$ then the possibilities are: 


*

*$\{A,B\}$

*$\{B,C\}$

*$\{C,A\}$



Each two-element set corresponds to $2 \times 1=2$ two-element ordered sets, since order matters in an ordered set, so while you have $3 \times 2 = 6$ possibilities putting on a shelf, you instead have $\dfrac{3 \times 2}{2 \times 1}=3$ possibilities putting in a bag
In terms of your post-scriptum, you might look at comparing 
A    1  1  -  2  2  - 
B    2  -  1  1  -  2 
C    -  2  2  -  1  1

with 
A    X  X  - 
B    X  -  X 
C    -  X  X

A: For the first question, I think you are correct. I can't be sure because the second paragraph is a little confusing, but to make things clear, each leaf in the tree you build is a possible arrangement of books on the shelf. For example, a few leaves might be $(4, 2, 7, 10)$, $(1, 2, 3, 4)$, and $(4, 3, 2, 1)$. Note that the last two are counted separately because the order matters.
For the second question, first we build the same tree as above. However, since we are just putting the books into a bag, it doesn't matter whether the order is $(1, 2, 3, 4)$ or $(4, 3, 2, 1)$: the same books are going in. Thus we have to remove 'duplicate' leaves, i.e. leaves which contain the same books but in different orders. For any set of books -- say $(a, b, c, d)$ where all books are distinct -- there are $4 \cdot 3 \cdot 2 \cdot 1 = 24$ ways to organize them by the previous reasoning. So each leaf in the original tree with $12 \cdot 11 \cdot 10 \cdot 9$ leaves is counted $4 \cdot 3 \cdot 2 \cdot 1$ times, meaning that the number of non-duplicate leaves is
$$\frac{12 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 3 \cdot 2 \cdot 1}$$
(Re. your PS: I'm not sure the 'bit-counting' method is very useful here, but the second question is equivalent to counting the number of 12-bit binary numbers with 4 '1's.)
