In how many ways can four distinct numbers be chosen from the set {0, 1, 2, ..., 9}, so that their sum is a multiple of 3?
There are: 10C4 =210 ways of choosing 4 distinct numbers. This includes combinations of 4 numbers that start with zero such as:0123, 0124, 0125.......etc.
Of the 210 combinations, the following have sums that are multiples of 3:
(123, 126, 129, 135, 138, 147, 156, 159, 168, 189, 234, 237, 246, 249, 258, 267, 279, 345, 348, 357, 369, 378, 456, 459, 468, 489, 567, 579, 678, 789, 1236, 1239, 1245, 1248, 1257, 1269, 1278, 1347, 1356, 1359, 1368, 1389, 1458, 1467, 1479, 1569, 1578, 1689, 2346, 2349, 2358, 2367, 2379, 2457, 2469, 2478, 2568, 2589, 2679, 3456, 3459, 3468, 3489, 3567, 3579, 3678, 3789, 4569, 4578, 4689, 5679, 6789) >>Total = 72 such sub-sets.
Note: The 30 or so numbers that have 3 digits are to be read, as I said, as: 0123, 0126...etc.
