11. (6 points) The average of 6 different non-zero natural numbers is 12. If one of the two-digit numbers $\overline{\mathrm{ab}}$ is replaced by $\overline{\mathrm{ba}}$ (where $a$ and $b$ are non-zero digits), then the average of these 6 numbers becomes 15. Therefore, the number of $\overline{\mathrm{ab}}$ that satisfy the condition is __.
【Solution】Solution: The original sum of the six numbers is: $12 \times 6=72$
The sum of the 6 numbers now is: $15 \times 6=90$
That is, $\overline{b a}-\overline{a b}=90-72=18$
Then, $10 b+a-10 a-b=18$
$$
\begin{array}{r}
9(b-a)=18 \\
b-a=2
\end{array}
$$

Since, the original sum of the six numbers is 72, the smallest sum of the other 5 numbers is $1+2+3+4+5=15$, then $\overline{\mathrm{ab}}$ is at most $72-15=57$, so $a$ is at most 5, then $a$ can be $5, 4, 3, 2, 1$, at this time $b$ is 7, 6, 5, 4, 3 respectively, $\overline{\mathrm{ba}}$ is at most $90-15=75$, so $b$ is at most 7, all the above situations are consistent.

Answer: There are 5 $\overline{\mathrm{ab}}$ that meet the conditions. The answer is: 5.