A Pandigital Divisible by Eleven

Write a nine-digit number using each of the digits $1$ to $9$ exactly once. What is the probability that it is divisible by $11$? A club of twenty-three once each wrote such a number, found two of them divisible by $11$, and declared the probability to be $\frac{2}{23}$. Were they right?

Solution

A number is divisible by $11$ when the sum of its digits in the odd positions and the sum in the even positions differ by a multiple of $11$. A nine-digit number has five odd positions and four even ones. Let $E$ be the sum of the four even-position digits; since all nine digits total $45$, the odd-position digits sum to $45 - E$. The test asks that $$(45 - E) - E = 45 - 2E$$ be a multiple of $11$, which happens exactly when $E$ leaves remainder $6$ on division by $11$.

The four even-position digits are distinct digits from $1$ to $9$, so $E$ lies between $1+2+3+4 = 10$ and $6+7+8+9 = 30$. The only values in that range leaving remainder $6$ are $E = 17$ and $E = 28$. Among the four-digit subsets of $\{1, \dots, 9\}$, nine sum to $17$ and two sum to $28$ (these two being $\{4,7,8,9\}$ and $\{5,6,8,9\}$), eleven choices in all.

For each such choice of which digits sit in the even positions, there are $4!$ ways to arrange them and $5!$ ways to arrange the other five. So out of all $9!$ arrangements, the proportion divisible by $11$ is $$\frac{11 \times 4! \times 5!}{9!} = \frac{11}{126} \approx 0.087.$$ The club’s $\frac{2}{23} \approx 0.087$ was a sample frequency that landed close to the truth by luck. The exact probability is $\frac{11}{126}$.