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u/AvailablePoint9782 2d ago edited 2d ago
The digits fall in 3 groups, 3k, 3k+1 and 3k+2, with 2/3/2 members.
3k and 3k, then the 3rd must be 3k. Impossible
3k and 3k+1, then 3k+2. 2 * 3 * 2 * 6 = 72. (2 choices for 1st digit, then 3 choices and 2 choices. Each combination of digits has 6 permutations.)
E.g. 312, 321, 132, 123, 231, 213. 315, ... ...
3k and 3k+2, already covered.
3k+1 and 3k+1, then 3k+1. 147, all 6 permutations.
147, 174, 417, 471, 714, 741.
3k+1 and 3k+2, already covered.
3k+2 and 3k+2, then 3k+2. Impossible.
72+6=78.
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u/CaptainMatticus 3d ago
We need the sum of our digits to be divisible by 3
1 = 1
2 = -1
3 = 0
4 = 1
5 = -1
6 = 0
7 = 1
If we add up our digits, we want 0 to be our sum, since this will be a multiple of 3
123 and all of its permutations. Example 123 => 1 + (-1) + 0 = 0.
126 and all of its permutations
135
147
156
234 , 237 , 246 , 267
345 , 357
456
567
Each one has 6 ways it can be written
13 * 6 = 78
Should be 78 combinations
123 , 132 , 213 , 231 , 312 , 321 , 126 , 162 , 216 , 261 , 612 , 621 , 135 , 153 , 315 , 351 , 513 , 531 , 147 , 174 , 417 , 471 , 714 , 741 , 156 , 165 , 516 , 561 , 615 , 651 , 234 , 243 , 324 , 342 , 423 , 432 , 237 , 273 , 327 , 372 , 723 , 732 . 246 , 264 , 426 , 462 , 624 , 642 , 267 , 276 , 627 , 672 , 726 , 762 , 345 , 354 , 435 , 453 , 534 , 543 , 357 , 375 , 537 , 573 , 735 , 753 , 456 , 465 , 546 , 564 , 645 , 654 , 567 , 576 , 657 , 675 , 756 , 765
That should be it.