Finding total number of possible numbers using below condition The questions was this
The number 916238457 is an example of a nine digit number which contains each of the digit 1 to 9 exactly once .It also has the property that digits 1 to 5 occur in their natural order while digits  1 to 6 do not .Find number of such numbers. 
So I did it like this
I first chose 5 places from 9 available places.This can be done in 9C5 ways.Then I arranged the remaining letters in 4! Ways .Now for the third condition if 6 would appear before 5 then it will be automatically satisfied.Now For 6 there are only two possibilities, it can either appear before 5 or after 5 so I divided my answer by 2!.
But my book says that the answer is 2520.
I don't know what I am missing .
Any help will be appreciated 
 A: Consider the string $12345$, the number $6$ can be inserted in $5$ places ( in order to fullfil the requirement of avoiding $123456$.) Now $7$ can be inserted in $7$ places,  $8$ can be inserted in $8$ places and $9$ can be inserted in $9$ places.
So that makes
\begin{eqnarray*}
5 \times 7 \times 8 \times 9 = \color{red}{2520} \text{  ways.}
\end{eqnarray*}
A: Its equal probability to see 1,2, 3, 4,5,6 in any of their all 720 ways to see them.
Now we want to see 612345 , 162345,126345,123645, or 123465.Which is 5 cases of 720 cases.
There are total of 9 factorial ways but we will multiply it by 5/720
9!*(5/720)=2520
A: Reword this till it is viable.
Lay out the $1$ to $5$  in order.  Before the $1$, between any two numbers, and after the $5$ are potential "pockets" where we can put more numbers. There are six such pockets.  
The $6$ is not in order so it can't go in the pocket after the $5$ and must go in one of the five pockets before the five but can go in any one of those $5$ pockets.
There are $5$ choices of where to put the six.  
Placing the $6$ in a pocket, splits the existing one pocket into two and we have a new pocket immediately before and a new pocket immediately after the $6$.  There as $7$ pockets now.
And we may place the seven in any of those $7$ pockets.  
The seven will split a pocket to make $8$ pockets to place the $8$.  And placing the eight in one of them will create $9$ pockets to place the $9$.
So there will be $5\cdot 7\cdot 8 \cdot 9=2520$ such numbers.
....
Alternatively there are $9$ spaces to put that $9$ digits.  There are ${9\choose 5}$ spots to reserve for the $1$ to $5$ (which must be put in order).  There are $4$ remaining spots for the remaining $4$ and $4! ways to arrange them.  Thus ${9\choose 5}4!= 6*7*8*9$.  
But we cant have the six in order.  By same logic there are ${9\choose 6}3!$ to have the $1$ to $6$ in order so there are ${9\choose 5}4!-${9\choose 6}3!= 6*7*8*9- 7*8*9 =5*7*8*9$.
....
Or there are $9!$ ways to place them.  There are $5!$ ways to to arrange the $1$ to $5$ in order.  There are $6!$ ways to arrange the $1$ to $6$ in order so $\frac {9!}{6!}$ ways to place the $1$ to $6$ in order, and $\frac {9!}{5!} -\frac {9!}{5!} = 6*7*8*9 - 7*8*9$ to place the $1$ to $5$ in order and the $6$ not.
