combinatorics - How many ways can the digits $2,3,4,5,6$ be arranged to get a number divisible by 11

How many ways can the digits $2,3,4,5,6$ be arranged to get a number divisible by $11$

I know that the sum of the permutations of the digits should be divisible by 11. Also, the total number of ways the digits can be arranged is $5! = 120$.

Answer

Hint. By the divisibility rule by $11$ we have to count the arrangements $d_1,d_2,d_3,d_4,d_5$ of the digits $2,3,4,5,6$ such that $d_1+d_3+d_5-(d_2+d_4)$ is divisible by $11$. Notice that
$$-2=2+3+4-(5+6)\leq d_1+d_3+d_5-(d_2+d_4)\leq 4+5+6-(2+3)=10$$

therefore we should have $d_1+d_3+d_5=d_2+d_4=\frac{2+3+4+5+6}{2}=10$.

In how many ways we can do that?