$1 \cdot 109$ Choose 3 numbers from $0,1,2, \cdots \cdots, 9$ such that their sum is an even number not less than 10. How many different ways are there to do this?
[Solution 1] Taking 3 numbers from 5 even numbers, there are $C_{5}^{3}=10$ different ways. Taking 1 number from 5 even numbers and 2 numbers from 5 odd numbers to form a triplet, there are $5 C_{5}^{2}=$ 50 different ways. Therefore, there are 60 different ways to get a sum that is even.

Among these 60 ways, the number of ways where the sum of the 3 numbers is less than 10 includes $\{0,1,3\}, \{0,1,5\}, \{0,1,7\}, \{0,2,4\}, \{0,2,6\}, \{0,3,5\}, \{1,2,3\}, \{1,2,5\}, \{1,3,4\}$, a total of 9 ways.
In summary, there are 51 different ways that meet the requirements.
[Solution 2] (1) Triplets containing 9 total 20 (taking 1 from the remaining 4 odd numbers and 1 from 5 even numbers);
(2) Triplets not containing 9 but containing 8 total 12;
(3) Triplets not containing 9 or 8 but containing 7 total 12, but $\{7,1,0\}$ is not valid;
(4) Triplets not containing 9, 8, or 7 but containing 6 include $\{6,5,3\}, \{6,5,1\}, \{6,4,2\}, \{6,4,0\}$, and $\{6,3,1\}$, a total of 5 groups;
(5) Triplets not containing 9, 8, 7, or 6 but containing 5 and having a sum of at least 10 include $\{5,4,3\}$, $\{5,4,1\}$, and $\{5,3,2\}$, a total of 3 groups.
In summary, there are $20+12+11+5+3=51$ different ways that meet the requirements.