4. (10 points) From the eight natural numbers 1 8, the number of ways to choose three numbers that include consecutive natural numbers is $\qquad$.

Take three numbers from the eight natural numbers 1 8, where there are consecutive natural numbers, the number of such combinations is $\qquad$.
【Analysis】The selection without consecutive natural numbers can be divided into four cases: (1) three odd numbers, (2) three even numbers, (3) two odd numbers and one even number, (4) two even numbers and one odd number; from the natural numbers $1 \sim 8$, selecting three numbers has a total of $\mathrm{C}_{8}^{3}$ $=56$ ways, then subtract the number of ways without consecutive natural numbers from 56 to solve.

【Solution】Solution: The number of ways without consecutive natural numbers: (1) three odd numbers: 135, 137, 157, 357, a total of 4;
(2) three even numbers: $246, 248, 268, 468$, a total of 4;
(3) two odd numbers and one even number: 136, 138, 158, 147, 358, 257, a total of 6;
(4) two even numbers and one odd number: $247, 258, 146, 148, 168, 368$, a total of 6;
In summary, the number of ways without consecutive natural numbers is: $4+4+6+6=20$ (ways)
The number of ways to select three numbers from $1 \sim 8$ is: $\mathrm{C}_{8}^{3}=\frac{8 \times 7 \times 6}{3 \times 2 \times 1}=56$ (ways)
Therefore, the number of ways with consecutive natural numbers is $56-20=36$ (ways);
Answer: From 1 to 8, selecting three numbers, the number of ways with consecutive natural numbers is 36.

The answer is: 36.