3. From the eight natural numbers $1 \sim 8$, the number of ways to choose three numbers such that there are no consecutive natural numbers is $\qquad$ kinds.
【Key Point】Counting
【Answer】 20
【Analysis】Solution one: Enumeration method
(1) Three odd numbers: 135, 137, 157, 357, 4;
(2) Three even numbers: $\underline{246}, \underline{248}, \underline{268}, \underline{468}, 4$;
(3) Two odd and one even: 136, 138, 158, 147, 358, 257, 6;
(4) Two even and one odd: $247, 258, 146, 148, 168, 368, 6$; in total $4+4+6+6=20$.
Solution two: Exclusion method
Choosing any three numbers from 1 to 8, there are $C_{8}^{3}=56$ different ways.
Among them, three consecutive numbers have 6 ways (123 $\underline{678}$), then the ways that meet the requirements are $56-6-30=20$.