Problem 690

51
3. 【Analysis and Solution】The number of ways to choose 3 different kilometer numbers from these 10 numbers is $\mathrm{C}_{5}^{3}$; the number of ways to choose 1 even number and 2 different odd numbers is $\mathrm{C}_{5}^{1} \mathrm{C}_{5}^{2}$.
The number of ways to choose 3 numbers from these 10 numbers such that their sum is less than 10 kilometers is as follows:
$$(0,1,3) ;(0,1,5),(0,2,4),(1,2,3),(0,1,7),(0,2,6),(0,3,5),(1,2,5),(1,3,4) \text {. }$$

Therefore, the number of different ways that meet the requirements of the problem is
$$C_{5}^{3}+C_{5}^{1} C_{5}^{2}-9=51 \text { (ways). }$$