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9 years, 5 months ago
9 years, 5 months ago
If the three digits are a, b and c.
$$a + b + c \leq 9$$ => a + b + c + k = 9, where k is a constant => $$^{12}C_3$$ cases = 220
But we need to remove the case where a, b and c are zero and k is 9 => 220 - 1 = 219
219 natural numbers have less than 4 digits and sum of digits less than 10.