Eleven

In this problem, we refer to the digits of a positive integer as the sequence of digits required to write it in base $10$ without leading zeros. For instance, the digits of $n$ = $2090$ are of course $2$, $0$, $9$ and $0$.

Let $n$ be a positive integer. We call a positive integer $m$ an eleven-multiple-anagram of $n$ if and only if

1. the digits of $m$ are a permutation of the digits of $n$, and

2. $m$ is a multiple of $11$.

You are required to write a program that given $n$, calculates the number of its eleven-multiple-anagrams.

As an example, consider again $n$ = $2090$. The values that meet the first condition above are $2009$, $2090$, $2900$, $9002$, $9020$ and $9200$. Among those, only $2090$ and $9020$ satisfy the second condition, so the answer for $n$ = $2090$ is $2$.