Number of words with 1 ahead of 2 Let A = {0, 1, 2, 3, .....9} be a set consisting of different digits. The number of ways in which a nine digit number can be made in which 1 and 2 are present and 1 is always ahead of 2 and repetition of digits is not allowed is $(8!*x)/2$. What is x?
My attempt: Let the one's place of the nine digit number be filled with 2. Then there are 8 ways in which 1 can be kept and the remaining positions can be filled in 8! ways. If the  tenth's place is filled with 2 then there are 7 ways in which 1 can be placed and 8! ways in which the other digits can be arranged. 
Proceeding in this way......
                 Total no. of ways=(1+2+3+4+5+6+7+8)*8! = 36*8!.

But, this gives an answer for x no were near the original answer of 65. What was wrong in my method?
 A: Forget about who is ahead of whom. If we leave out $0$, there are $9!$ numbers. If we leave out any of the other $7$ digits that can be left out, we have $8\cdot 8!$ possibilities, since $0$ cannot be the first digit. The total is $9\cdot 8! +(7)(8)8!$, which is $(65)(8!)$.
In half of these, $1$ is ahead of $2$ and in half of them it is behind $2$.
A: Hint : for each $i<j$ let us denote by
$N_{i,j}$ the number of solutions where the $2$ is on the $i$-th
place and the $1$ is in the $j$-th place.
Then $N_{ij}$ is independent of $i,j$, it is always equal to [fill the blank].
A: $0$ must not be the leading digit, so we split into two cases, depending on whether or not $0$ is one of the digits. For the moment, let's not worry about which order $1$ and $2$ are put in. If $0$ is not one of the digits, then each other number in the set is a digit. How many ways are there to arrange these digits? If $0$ is a digit, then there are $7$ ways to choose the number in the set that isn't a digit. How many ways are there to arrange the chosen digits, since $0$ can't come first? Thus, how many $9$-digit numbers containing $1$ and $2$ with no repeated digits are there. Observing that $1$ comes ahead of $2$ half the time, then dividing by $2$ will get us the desired number, which will allow us to solve for $x$.
