9. From the 10 numbers $0,1,2,3,4,5,6,7,8,9$, choose 3 numbers such that their sum is an even number not less than 10. The number of different ways to do this is $\qquad$.


Since the sum of 3 numbers is an even number not less than 10, then these 3 numbers are either 3 even or 2 odd 1 even. When these 3 numbers are 3 even, except for $0,2,4$ and $0,2,6$, there are 8 cases that meet the condition; when these 3 numbers are 2 odd 1 even, if 0 is the chosen even number, then except for 1,$3 ; 1,5 ; 1,7 ; 3,5$, there are 6 cases that meet the condition; if 2 is the chosen even number, then except for 1,$3 ; 1,5$, there are 8 cases that meet the condition; if 4 is the chosen even number, then except for 1,3, there are 9 cases that meet the condition; if 6,8 are the chosen even numbers, then there are 10 cases each that meet the condition. Therefore, there are $8+6+8+9+20=51$ different ways.