How many $7$ digits number can be made? How many $7$ digits number can be made with $1,2,3,4,5,6,7$ so that they are divisible by $11$? (Repetition is not allowed.) 
I know the divisibility rule of $11$, so the main problem is counting.
 A: Hint: work out a step by step procedure for choosing a number of the required type.  Think of the number as $ABABABA$.  Since all the digits add up to ... , the divisibility-by-$11$ rule says that all the $B$s must add up to ... (fill in the dots for yourself).
(1) First choose the three digits $B$ so that they add up to ... - there are not many possibilities and they are easy to count by trial and error.
(2) There is now only one choice for the $A$s, they are just the leftover numbers.
(3) Decide on an order for the $B$s.
(4) Decide on an order for the $A$s.
Work out the number of ways of doing steps $1,3$ and $4$ - there is only one way to do step $2$ so it doesn't matter - then combine them to get your final answer.
Good luck!
Update.  As pointed out by @chubakueno, there is "in principle" more than one option for the sum of the $B$s.  However only one of these options can be made the sum of three numbers from $1,2,3,4,5,6,7$.
A: If your number is $\overline{abcdefg}$, then $(b+d+f) - (a + c + e+g) = 0 \pmod{11}$. 
The difference is $0$ (I don't think $11$ or $-11$ is possible, $-11$ isn't possible because of parity).
For $0$ you can try the solutions for $14-14$, any other combination ends up with the left side too large or small. 
$(b+d+f)$ is one of these 4 combinations: $(2, 5, 7), \ (3, 5,6), \ (3,7,4), \ (1,7,6)$. 
Multiply 4 by how many orderings of 3 digits exist, then multiply by how many orderings of the last 4 digits exist $(4 \times 3! \times 4!)$
There are exactly 576 solutions. 
