<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] Since the raw digit sum of $1..9$ will be $45$, an odd number, we clearly can't have two equal sums for the even-position and odd-position digits. So we need to have one set of digits sum to $17$ and the others to $28$, so that the difference is divisible by $11$.
This can be done either way around; $28$ as the sum of the four even-position digits or the sum of the five odd-position digits.
Making 28 with only four digits is the more restrictive case. The only options are {9,8,7,4} and {9,8,6,5}. With five digits there are more choices: {9,8,7,3,1}, {9,8,6,4,1}, {9,8,6,3,2}, {9,8,5,4,2}, {9,7,6,5,1}, {9,7,6,4,2}, {9,7,5,4,3}, {8,7,6,5,2}, {8,7,6,4,3}.
So we have 11 ways of dividing the digits suitably, and in each case we can permute the five odd-position digits freely and similarly the four even-position digits. So the number of possibilities is:
$$11\cdot 5!\cdot 4! = 11\times 120\times 24 = 31680$$
As a sanity check, this total is reasonably close to $\frac{9!}{11}$, ie. one-eleventh of the total permutations of the digits.
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