We are tossing $20$ dice. Let $X_i$ be the number showing on the $i$-th die. We are interested in the random variable $X_1+\cdots+X_{20}$.
The $X_i$ are independent, mean $\frac{7}{2}$, variance $\frac{35}{12}$ (please verify).
So $Y$ has mean $\mu=20\cdot\dfrac{7}{2}$, variance $\sigma^2=20\cdot \dfrac{35}{12}$.
We cross our fingers and use the normal approximation. This is because $Y$ is the sum of a not terribly small number of independent identically distributed respectable random variables $X_i$.
So we want the probability that if $W$ is normal with mean $\mu$ and variance $\sigma^2$, then $30\le W\le 40$.
The rest depends to some degree on whether you are expected to use the continuity correction.
Without the continuity correction, we want the probability that a standard normal $Z$ satisfies $\frac{30-\mu}{\sigma}\le Z\le \frac{40-\mu}{\sigma}$.
With continuity correction, replace $40$ by $40.5$, and $35$ by $34.5$.
If you have any difficulty finishing, please leave a comment.
In your computation of how many ways for two dice to sum to 9, you subtracted 2 for the two cases in which $x_1$ is 7 or 8, but forgot to subtract 2 for the two cases in which $x_2$ is 7 or 8.
Using a formula is nice, but especially when what's going on is as simple as two numbers from 1 to 6 summing to 9, double-check it by writing them out!
Best Answer
Here's another approach to counting the numerator:
Take the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9.$ Arrange them in any sequence. Then pick three of the numbers $1, 2, 3, 4, 5, 6$ and replace $7, 8, 9$ with those numbers in increasing order (that is, $7$ is replaced by the smallest number, $9$ by the largest).
This produces every possible sequence of nine numbers selected from $\{1, 2, 3, 4, 5, 6\}$ with three pairs and three singletons, that is, the number of rolls that have exactly three pairs and no other matches, but it produces each such sequence more than once. In particular, each of the numbers $7, 8,$ and $9$ could originally have appeared either before or after the numbers they eventually are made to duplicate, so each roll of three pairs is produced exactly eight times.
After correcting for this overcounting, the number of rolls is
$$ \frac{9! \binom63}{8} = \frac52(9!).$$
This turns out to be equal to
$$ \binom{6}{3}\binom{9}{2}\binom{7}{2}\binom{5}{2}\times 6 = \binom{6}{3} \frac{9!7!5!\times6}{(2!7!)(2!5!)(2!3!)} = \binom{6}{3} \frac{9!}{(2!)^3},$$
confirming your answer for the probability of exactly three pairs.