7. In how many ways can the digits 1, 2, 4, 7, and 9 be arranged to form a five-digit number divisible by 11?
(A) 0
(B) 1
(C) 10
(D) 12
(E) 24.
7. The answer is (D). A number is divisible by 11 if and only if the difference between the sum of the digits in odd positions and the sum of the digits in even positions is divisible by 11.

Let $a$ be the sum of the 3 digits in odd positions and $b$ be the sum of the 2 digits in even positions. Since $a+b=1+2+4+7+9=23$ is odd, $a-b$ will also be odd and therefore different from 0. Furthermore, since $b$ is at most $7+9=16$ and at least $1+2=3$, it follows that $a-b=(23-b)-b=23-2b$ is between -9 and 17. If $a-b$ must be divisible by 11, the only possibility is that it is 11, but then $11=a-b=23-2b$ from which $b=6$. The only pair of digits among those considered that gives a sum of 6 is $\{2,4\}$.

In conclusion, a multiple of 11 is obtained if and only if the digits in even positions are 2 and 4. In total, there are 12 combinations because there are 2 ways to fix the order of the digits in even positions and $3!=6$ ways to fix the order of the other 3.