3. (10 points) From the eight natural numbers $1 \sim 8$, choose any three numbers, such that there are no consecutive natural numbers in the selection. The number of ways to do this is.
20
【Analysis】First, taking 3 out of all has $\mathrm{C}_{8}^{3}=56$ ways.
If two of the numbers are consecutive and the other one is not, if these two consecutive numbers are at the ends, which are 12 or 78, then each has 5 different ways,

such as: 124, 125, 126, 127, 128,
If these two numbers are in the middle, which are $23, 34, 45, 56, 67$, then each has 4 different ways,
such as: 235, 236, 237, 238;
In total, there are $5 \times 2+5 \times 4$ ways;
Three consecutive numbers have 6 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 pick without consecutive natural numbers is 20.
Therefore, the answer is: 20.