Answer
30,492 positive integers less than 1,000,000 have their sum equal to 19.
Work Step by Step
A number less than 1,000,000 will have at most 6 digits. Let those digits be $x_{1},x_{2},x_{3},x_{4},x_{5},x_{6}$
$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}$ = 19
And 0$\leq x_{i}\leq$9
Ways to select 19 distinguishable objects from 6 distinguishable boxes= $C$(24,19) = 42,504
Number of solutions with $x_{i}\geq$10
Let us define $X_{i}$ = $x_{i}$ - 10
So for 1st digit we have
$X_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}$ = 9
$C$(14,9)= 2002
We can only have 1 $X_{i}$ because 2$X_{i}$s will make the sum greater than 19.
Since $X_{i}$can take 6 values we multiply 2002 by 6 i.e. 12,012.
So 42,504 solutions are there without restrictions, but with the restriction, we have only 42,504-12,012 = 30,492 solutions
