3. (10 points) From the eight natural numbers 1 $\sim$ 8, if we randomly select three numbers, the number of ways to select them such that there are no consecutive natural numbers is $\qquad$ kinds.
3. (10 points) From the eight natural numbers 1 to 8, choose any three numbers such that there are no consecutive natural numbers. The number of ways to do this is $\qquad$.

【Analysis】First, the total number of ways to choose 3 numbers from 8 is $C_{8}^{3}=56$.
If two of the numbers are consecutive and the other one is not, and if these two consecutive numbers are at the ends, such as 12 or 78, there are 5 different methods for each,

For example: 124, 125, 126, 127, 128,
If these two consecutive numbers are in the middle, such as $23, 34, 45, 56, 67$, there are 4 different methods for each,
For example: 235, 236, 237, 238;
In total, there are $5 \times 2+5 \times 4$ methods;
If there are three consecutive numbers, there are 6 possible cases: $123, 234, 345, 456, 567, 678$;
Subtract the number of ways with consecutive natural numbers from the total number of ways to get the number of ways that meet the requirement.
【Solution】Solution: $\mathrm{C}_{8}^{3}=\frac{8 \times 7 \times 6}{3 \times 2 \times 1}=56$ (ways)
The number of ways with two consecutive numbers is:
$5 \times 2+5 \times 4=30$ (ways)
The number of ways with three consecutive numbers is 6:
$56-30-6=20$ (ways)
Answer: The number of ways to choose three numbers with no consecutive natural numbers is 20.
Therefore, the answer is: 20.