3. (10 points) From the eight natural numbers 1 to 8, choose any three numbers such that there are no consecutive natural numbers among them. There are $\qquad$ ways to do this.
【Analysis】First, the number of ways to choose 3 out of all is $\mathrm{C}_{8}^{3}=56$.
For two consecutive numbers and another one not consecutive, if these two consecutive numbers are at the ends, which are 12 or 78, there are 5 different ways for each,

such as: 124, 125, 126, 127, 128,
If these two numbers are in the middle, which are $23, 34, 45, 56, 67$, there are 4 different ways for each, such as: 235, 236, 237, 238;

In total, there are $5 \times 2+5 \times 4$ ways;
For 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, and you 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 possible ways with two consecutive numbers are:
$$
5 \times 2+5 \times 4=30 \text { (ways) }
$$

The possible ways with three consecutive numbers are 6:
$$
56-30-6=20 \text { (ways) }
$$

Answer: The number of ways to choose without consecutive natural numbers is 20.
Therefore, the answer is: 20.