Taking advantage of the fact that thepartitions have only 3 elements.
If the first number is i, the second number can be anything from 1 to
999999-iand the third number is then uniquely fixed.
Since ican run from 1 to 999998, the total number of such choices is (-:
999998 * 999999).
But this counts triplets more than once. Any triplet in which the
numbers are unique will be counted 6 times.
Any triplet containing a repetition will be counted 3 times: once when i
is the unrepeated number and twice when i is the repeated number. This
happens here whenever i is even, thus
(-:999998) times.
Any triplet containing all 3 numbers equal will be counted just once,
but there aren't any of them.
To find the unique triplets, we take (-: 999998 * 999999), add in 3 *
(-:999998) to bring the 3-times-repeated cases up to 6-times-repeated,
then divide by 6:
0 ": 6 %~ (-: 999998 * 999999) + 3 * -: 999998
83333333333
Henry Rich
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