In how many ways can three distinct numbers be chosen from the set {1, 2, 3, ..., 9}, so that their sum is a multiple of 3?
+23245 Start listing them (have the second digit larger than the first and the third digit larger than the second):
123 126 129 135 138 147 156 159 168
234 237 246 249 258 267 279
345 348 357 369 378
456 459 468 489
567 579
678
789
But each of these can be chosen in 6 ways: 123 = 132 = 213 = 231 = 312 = 321
So take the number of ways listed and multiply by 6.
It would also be wise to check to see if I have missed any.
