An encyclopedia consists of 8 volumes, numbered 1 to 8, initially arranged in ascending order of their numbers An encyclopedia consists of 8 volumes, numbered 1 to 8, initially arranged in ascending order of their numbers.
a) How many ways can we put these volumes on a shelf?
b) In how many ways can we put these volumes on a shelf so that at least one volume is not occupying the same starting position?
c) In how many ways can we place these volumes on a shelf so that exactly one volume is not placed in ascending order? For example, 2, 1, 3, 4, 5, 6, 7, 8 or 1, 2, 4, 5, 3, 6, 7, 8.
d) In how many ways can we select 3 of the 8 volumes so that they do not have consecutive numbers?
Book answers:
a) $8!$ b) $8!-1$ c) $8\cdot7 - 7$ or $7\cdot7$ d) $\text{C}_6^3$
My answers:
a) Permutation of 8 different volumes in a row: $8!$ (correct)
b) I fixed the 1 in first place and changed the other 7, because within the possibilities, only the 1 can be in its initial place with all out of their respective initial places, or all can match in their respective initial places as well, as the utterance asks for at least 1 in the starting place then it would, by indirect count: $8! −7!$ (but incorrect, I don't understand why it subtracts only one possibility if one of them would be in the case that they all coincided in the starting position, but it doesn't necessarily have to happen, as the statement says at least one)
c) As it's just one number out of place, keeping everyone in their starting position and choosing 1 of the 8 to put between them we have 7 possibilities to allocate with 8 possible numbers: $8\cdot7$ (but incorrect, I don't understand why it subtracts 7 from this count)
d) As it must not be the 3 consecutive ones, for one of the numbers in the count we cannot count its smallest consecutive and its biggest consecutive must not be included in the count, so as there are 8 numbers we will only have 6 to choose 3: $\text{C}_6^3$, but as we need to choose 6 out of 8 totals possibles so getting $\text{C}_6^3\cdot\text{C}_8^6$ (but also incorrect)
What did I think wrong and how should I have thought?
 A: $\mathbf{\text{ a-)}}$ It is correct , as you write.
$\mathbf{\text{ b-)}}$ All arrangements are $8!$ , but we do not want the only position which is thenumbers are in  their original placement like $(1,2,3,4,5,6,7,8)$ etc., so $8! -1$ where $1$ represent their starting arrangements , it is trivial
$\mathbf{\text{ c-)}}$ According to given answer , i tihnk that your book tought that when you place the selected number any of suitable $7$ position , then it will cause another non-ascending situation .For example , for $(1,2,3,4,8,5,6,7)$ , $8$ will give rise to another non-asceding number such as $5$ , so they subtract $7$ to eliminate the numbers like $5$ in this example
$\mathbf{\text{EDIT:}}$ It seems that the second part of part $c$  has misunderstanding , thanks for this to @N.F.Taussig. As i wrote , we have $8$ choices to select a number and $7$ places to put it , so $8 \times 7 =56$ .However , as @N.F.Taussig mentioned , when we do that , we also count the changing order of adjacent numbers twice.Because if we take $7$ and put it in front of $6$ , then it also count that $6$ is taken and placed beyond of $7$. Hence , we should subtract the number of adjacent pairs.As a result , $56-7=49$
$\mathbf{\text{ d-)}}$  Lets say that our selected number denoted by $a$ and the rest by $b$ , then we can say that  , we want to places $a's$ so that they are not adjacent in given line such that $$-b-b-b-b-b-$$ , we can do it by $C(6,3)=20 $ ways. For example , one of the arrangements is $a,b,b,a,b,a,b,b =(1,2,3,4,5,6,7,8,9)$ , then the selected numbers are $(1,4,6)$
A: An encyclopedia consists of 8 volumes, numbered 1 to 8, initially arranged in ascending order of their numbers.
a) How many ways can we put these volumes on a shelf?
You are correct that the eight volumes can be permuted in $8!$ ways.
b) In how many ways can we put these volumes on a shelf so that at least one volume is not occupying the same starting position?
At least one volume does not occupy the same starting position unless all the volumes are placed in their original positions.  The only permutation that leaves all volumes in their original position is the identity permutation $(1, 2, 3, 4, 5, 6, 7, 8) \to (1, 2, 3, 4, 5, 6, 7, 8)$, so there are $8! - 1$ permutations that leave at least one volume not in its starting position.
c) In how many ways can we place these volumes on a shelf so that exactly one volume is not placed in ascending order? For example, 2, 1, 3, 4, 5, 6, 7, 8 or 1, 2, 4, 5, 3, 6, 7, 8.
There are eight ways to select the volume that is out of place and seven other positions to which that volume can be moved.  For instance, if we select $3$ and move it to the fifth position, we obtain the sequence $(1, 2, 4, 5, 3, 6, 7, 8)$ since the remaining seven numbers must be placed in ascending order.  This suggests that there are $8 \cdot 7$ sequences in which exactly one volume does not appear in ascending order.
However, we have counted each case in which we switched the positions of two adjacent numbers twice, once when we moved the larger number to the smaller number's position and once when we moved the smaller number to larger number's position. For instance, if we move $1$ to the second position, we obtain the sequence $(2, 1, 3, 4, 5, 6, 7, 8)$.  We also obtain the sequence $(2, 1, 3, 4, 5, 6, 7, 8)$ if we move $2$ to the first position.  There are seven pairs of adjacent numbers.
Therefore, there are $8 \cdot 7 - 7 = 7 \cdot 7$ sequences in which exactly one volume does not appear in ascending order.
d) In how many ways can we select 3 of the 8 volumes so that they do not have consecutive numbers?
This is equivalent to asking in how many ways we can line up five blue and three green balls so that no two of the green balls are consecutive.  Line up the five blue balls.

Doing so creates six spaces in which to place the green balls, four between successive blue balls and two at the ends of the row.  To ensure that no two of the green balls are consecutive, we must choose three of these six spaces in which to place one green ball each, which can be done in $\binom{6}{3}$ ways.  For instance, if we select the third, fourth, and sixth spaces, we obtain the arrangement shown below.

Now number the balls from left to right.  The numbers on the green balls are
the numbers of the volumes on the selected encyclopedias, no two of which are consecutive.
