In how many ways can three distinct numbers be chosen from the set {1, 2, …, 100} so that their sum is even?

```
Sum is even when all the three integers are even or when two are odd and the third is even. So the answer is (50 C 3) + 50*(50 C 2) = 80850.
Here (50 C 3) means combination of 50 objects taken 3 at a time.
[b]This C code on running gives an output(number of ways) as 80850.[/b]
#include
int main(){
int i, j, k;
int count = 0;
for(i = 1; i <= 100; i++){
for(j = 1; j <= 100; j++){
for(k = 1; k <= 100; k++){
if((i+j+k)%2==0 && i!=j && j!=k && k!=i)
count++;
}
}
}
printf("%d", count/6);//count to be divided by 6(3!)
return 0;
}
```