I came up with 2,872.
I don't have an elegant proof or anything, though. I just methodically listed the possible combinations, and then used some combinatorics to total up the number of permutations. The following shows the different combinations, followed by the number of permutations for each (e.g. 13 can be made from three 4s and one 1, and there are four different ways to order this sum).
(4, 4, 4, 1) - 4
(4, 4, 3, 2) - 12
(4, 4, 3, 1, 1) - 30
(4, 4, 2, 2, 1) - 30
(4, 4, 2, 1, 1, 1) - 60
(4, 4, 1, 1, 1, 1, 1) - 21
(4, 3, 3, 3) - 4
(4, 3, 3, 2, 1) - 60
(4, 3, 3, 1, 1, 1) - 60
(4, 3, 2, 2, 2) - 20
(4, 3, 2, 2, 1, 1) - 180
(4, 3, 2, 1, 1, 1, 1) - 210
(4, 3, 1, 1, 1, 1, 1, 1) - 56
(4, 2, 2, 2, 2, 1) - 30
(4, 2, 2, 2, 1, 1, 1) - 140
(4, 2, 2, 1, 1, 1, 1, 1) - 168
(4, 2, 1, 1, 1, 1, 1, 1, 1) - 72
(4, 1, 1, 1, 1, 1, 1, 1, 1, 1) - 10
(3, 3, 3, 3, 1) - 5
(3, 3, 3, 2, 2) - 10
(3, 3, 3, 2, 1, 1) - 60
(3, 3, 3, 1, 1, 1, 1) - 35
(3, 3, 2, 2, 2, 1) - 60
(3, 3, 2, 2, 1, 1, 1) - 210
(3, 3, 2, 1, 1, 1, 1, 1) - 168
(3, 3, 1, 1, 1, 1, 1, 1, 1) - 36
(3, 2, 2, 2, 2, 2) - 6
(3, 2, 2, 2, 2, 1, 1) - 105
(3, 2, 2, 2, 1, 1, 1, 1) - 280
(3, 2, 2, 1, 1, 1, 1, 1, 1) - 252
(3, 2, 1, 1, 1, 1, 1, 1, 1, 1) - 90
(3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) - 11
(2, 2, 2, 2, 2, 2, 1) - 7
(2, 2, 2, 2, 2, 1, 1, 1) - 56
(2, 2, 2, 2, 1, 1, 1, 1, 1) - 126
(2, 2, 2, 1, 1, 1, 1, 1, 1, 1) - 120
(2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1) - 55
(2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) - 12
(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) - 1
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Posted by tomarken
on 2014-07-21 16:31:11 |
Solution
