Dominoes Again

Source: Original

Solution received from Jon Abraham:

The pair set aside must have a number of spots X such that the quantity 50-X is divisible by four. A Sum=10 solution, setting aside the 2-2 and the 3-3, and a Sum=9 solution, setting aside the 3-3 and 4-4. A Sum=11 solution can be found by substituting (5-i) for every (i) in the Sum=9.

4-1 3-2 1-1 4-4 3-4 2-1 2-4 1-3 |
2-2 2-3 1-2 2-4 3-1 4-1 3-4 1-1 |

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