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 Getting 13 (Posted on 2014-07-21)
Using integers 1,2,3 & 4 only , in how many distinct ways a sum of 13 can be achieved?

Rem: The order matters i.e. 1,4,4; 4,1,4; & 4,4,1 are considered distinct.

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 Solution | Comment 1 of 5

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

 Posted by tomarken on 2014-07-21 16:31:11

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