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 $5! = 120$.
 A: 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? 
A: Let the base-$10$ number $abcde$ be divisible by $11,$ where  $\{a,b,c,d,e\}=\{2,3,4,5,6\}.$ The base-$10$ number $abcde$ is divisible by $11$ iff $11$ divides $(a+c+e)-(b+d).$ 
We cannot have $a+c+e\geq 11 +b+d$ because $a+c+e\leq 4+5+6=15$ and $11+b+d\geq 11+2+3=16.$ 
We cannot have $b+d\geq 11+a+c+e$ because $b+d\leq 5+6=11$ and $11+a+c+e\geq 11+2+3+4=20.$ 
So the only multiple of $11$ that can be equal to  $(a+c+e)-(b+d)$ is $0.$  So $a+c+e=b+d.$ Hence $2(b+d)=(a+c+e)+(b+d)=20.$ 
So $b+d=10.$ 
So $6\in \{b,d\}.$ Otherwise $b+d\leq 5+4=9$, but $b+d=10$. Since $6\in \{b,d\}$ and $b+d=10$ we have $\{b,d\}=\{6,4\}.$ Hence $\{a,c,e\}=\{2,3,5\}.$ 
There are $3!$ permutations of $(2,3,5)$ and $2!$ permutations of $(4,6)$ so there are $3!\cdot 2!=12$ ways.
A: The total sum of all digits given is $20$. Hence we need to break it into two equal parts which is $10$ each.  And the only pair forming it using given digits is {2,3,5}, {4,6}
We know the three digits can be placed only on $1^{st},3^{rd}$ and $5^{th}$  places respectively while the rest on remaining.  Hence the number of ways is simply permitting 3 numbers and 2 numbers at the same time. 
Hence the answer would be simply $3!*2!=12$
