<s>[INST] How many ways a 9 digit number can be formed using the digits 1 t0 9 without repetition such that it is divisble by $11$. How many ways a 9 digit number can be formed using the digits 1 t0 9 without repetition such that it is divisible by $11$.
My attempt-
A number is divisible by 11 if the alternating sum of its digit is divisible by 11?
The other thing to notice is as it is a 9 digit number formed by digits 1 to 9, exactly once each digit from 1 to 9 will appear in the number. 
Basically, the question boils down to how many ways we can arrange 123456789 so that the alternating sum of the digit is divisible by 11.
I am not able to proceed further. Any help would be appreciated.
 [/INST] $$1+2+...+9=\frac{9\cdot10}{2}=45\\(x_1+x_3+x_5+x_7+x_9)-(x_2+x_4+x_6+x_8)=11m$$
$m$ must be $1$ or $3$ since 45 is odd and $3$ is discarded immediately so we have 
$$\begin{cases}X+Y=45\\X-Y=11\end{cases}$$ Hence $X=28$ and $Y=17$ We work with $28$.
1) With $9$ and $8$ one has $x_1+x_2=11$ and $x_1+x_2+x_3=11$ (because $28-(9+8)=11$).
$x_1+x_2=7+4=6+5$ give $2$ cases.
$x_1+x_2+x_3=7+3+1=6+4+1=6+3+2=5+4+2$ give $4$ cases.
2) With $9$ and $7$ without $8$ one has  $x_1+x_2=12$ and $x_1+x_2+x_3=12$.
$x_1+x_2=12$ no cases since $6+5=11\lt12$.
$x_1+x_2+x_3=6+5+1=6+4+2=5+4+3$ give $3$ cases.
With $9$ and $6$ without $7$ and $8$ is not possible because $9+6=15$ and $5+4=9\lt 13$.
3) With $8$ and $7$ without $9$ one has $x_1+x_2$ not possible and
$x_1+x_2+x_3=6+5+2=6+4+3$ give $2$ cases.
Hence there are $2+4+3+2=11$ possible cases each of these corresponding to permutations of $5$ and $4$ digits.
Thus there are $11\cdot4!\cdot5!=\color{red}{31680}$ possibilities.
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