Is this claim correct? 
The digits $2,3,4,7$ and $8$ will be put in random order to make a positive five-digit integer. What is the probability that the resulting integer will be divisible by $11$?

I got the correct answer to this problem by using the rule that if the alternating sum of the digits of a number is divisible by $11$, then so is that number. However, the official solution did it some other way which I believe is invalid. It is solved using the claim that a number is divisible by $11$ if the sum of every other digit starting with the first one equals the sum of every other digit starting with the second one. This can easily be disproved by looking at $2728$, for example. Here, $2+2 \neq 7+8$ yet $2728$ is divisible by $11$. Using this false claim the solution got the correct answer of $\dfrac{1}{10}$, but I don't know if it was a coincidence or something else.
Here is the solution the book used for reference:
 A: Let the $5$ digit number be $N = a_1a_2a_3a_4a_5$. 
The correct divisibility test is that $N$ is divisible by $11$ iff $a_1-a_2+a_3-a_4+a_5$ is divisible by $11$, i.e. $a_1-a_2+a_3-a_4+a_5 \equiv 0 \pmod{11}$, i.e. $a_1+a_3+a_5 \equiv a_2+a_4 \pmod{11}$. 
They covered the case where $a_1+a_3+a_5 = a_2+a_4$, but they forgot to account for the possibility that $a_1+a_3+a_5$ and $a_2+a_4$ differ by some non-zero multiple of $11$. 
For this problem, since the sum of $2,3,4,7,8$ is even, we know that $a_1+a_3+a_5$ and $a_2+a_4$ will differ by an even number.
Also, $9 = 2+3+4 \le a_1+a_3+a_5 \le 4+7+8 = 19$ and $5 = 2+3 \le a_2+a_4 \le 7+8 = 15$. Hence, $a_1+a_3+a_5$ and $a_2+a_4$ can't differ by more than $14$. 
Thus, it is impossible for $a_1+a_3+a_5$ and $a_2+a_4$ to differ by a non-zero multiple of $11$. This is why the book got the same answer you did, even though they forgot to cover a few cases.
A: Your reasoning is correct. The book was incorrect, but luckily for them the digits that they gave you did not allow for any possibilities like the $2728$ that you have described. Th should have considered poissibilities where the sum of the three digits were 11 more than the sum of the other two digits and others.
A: The book's claim is true. If the two sums they describe are equal, then the alternating sum you describe is $0,$ which is divisible by $11,$ and so the number is divisible by $11$.
Still, it would be better to have used a result equivalent to divisibility by $11$, rather than one that is sufficient (but not necessary).
A: I think that the book-answer statement is indeed missing the fact that the alternating sums have to be equal mod 11 -- which I also think is equivalent to your alternative statement. For your example of $2728$, note that $2 + 2 = 4 = 15 \mod 11 = (7 + 8) \mod 11$. 
