The total number of 3-digit numbers, whose sum of digits is 10, is __________.

Solution

Maths sir · 4 steps

Shift the first digit down by 1, count with stars and bars, then remove the one bad case.

1. Set up digits. Call the digits a, b, c with a + b + c = 10. The hundreds digit a cannot be 0, so 1 ≤ a ≤ 9, and 0 ≤ b, c ≤ 9.
2. Remove the lower limit. Put a' = a - 1, so a' ≥ 0 and the equation becomes a' + b + c = 9. Now all three unknowns are non-negative integers.

   Solutions of a' + b + c = 9 = C(9 + 3 - 1, 3 - 1) = C(11, 2) = 55

3. Check upper limits. Since b + c ≤ 9, neither b nor c can go above 9. But a' = 9, b = c = 0 gives a = 10, which is not a digit. That is the only bad case.
4. So. The count is 55 - 1 = 54, so there are 54 such numbers.

The common slip: 55. Forgetting that the hundreds digit cannot be 10 leaves the case (10, 0, 0) in the count and gives 55 instead of 54.

The idea to keep. For digit sums, shift any digit that has a lower limit, use C(n + r - 1, r - 1), then check whether any digit can go above 9 and subtract those cases.

Answer from the NTA final answer key. Solution by PrepWiser.