Solution:

The answer is (D). A number is divisible by $11$ if and only if the difference between the sum of the digits in odd position and the sum of the digits in even position is divisible by $11$.

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

In conclusion, one obtains a multiple of $11$ every time, and only when, the digits in even position are $2$ and $4$. In total there are $12$ combinations, because there are $2$ ways to fix the order of the digits in even position and $3!=6$ ways to fix the order of the other $3$.