Converting each number to base 10 \(110_{two} = 1 \times 2^2 + 1 \times 2^1 + 0 \times 2^0\) = 1 \times 4 + 1 \times 2 + 0 \times 1 = 4 + 2 + 0 = \(6_{ten}\) \(31_{eight} = 3 \times 8^1 + 1 \times 8^0\) = 3 \times 8 + 1 \times 1 = 24 + 1 = \(25_{ten}\) \(42_{five} = 4 \times 5^1 + 2 \times 5^0\) = 4 \times 5 + 2 \times 1 = 20 + 2 = \(22_{ten}\) Hence, \(31_{eight} > 42_{five} > 110_{two}\) In ascending order, \(110_{two}, 42_{five}, 31_{eight}\)