We need to count 3-digit integers (100-999) whose digit sum equals 14.
Let the three-digit number have digits $$a$$ (hundreds), $$b$$ (tens), $$c$$ (units) where $$1 \leq a \leq 9$$, $$0 \leq b \leq 9$$, $$0 \leq c \leq 9$$, and $$a + b + c = 14$$.
Since $$a \geq 1$$, set $$a' = a - 1$$ so that $$0 \leq a' \leq 8$$ and the equation becomes $$a' + b + c = 13$$.
Without considering the upper bounds on $$a'$$, $$b$$, and $$c$$, the number of nonnegative integer solutions is given by stars and bars as $$\binom{13 + 2}{2} = \binom{15}{2} = 105$$.
To account for the constraints, subtract cases where one of the variables exceeds its bound using inclusion-exclusion.
If $$a' \geq 9$$, let $$a'' = a' - 9$$, giving $$a'' + b + c = 4$$ and $$\binom{4 + 2}{2} = 15$$ solutions. If $$b \geq 10$$, let $$b' = b - 10$$, giving $$a' + b' + c = 3$$ and $$\binom{3 + 2}{2} = 10$$ solutions. If $$c \geq 10$$, let $$c' = c - 10$$, giving $$a' + b + c' = 3$$ and $$\binom{3 + 2}{2} = 10$$ solutions. Any intersection of two or more such violations is impossible because it would require a sum exceeding 13.
Hence the total count is $$\text{Count} = 105 - 15 - 10 - 10 = 70$$.
The answer is 70.
91- 6303239042
