How many positive integers $<1,000,000,000$ have exactly one digit equal to $9$ and have the sum of digits equal to $13$?
I have:
$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + 9 = 13$
$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 = 4$
Then I use ${{n+k-1}\choose k}$:
${11\choose 4}$ = 330
Then $330*9 = 2970$.
So there are $2970$ integers that satisfy the problem.
My questions:
First, is this correct?
Second, why/when should I use ${{n+k-1}\choose k}$ over ${{n+k-1}\choose k-1}$
Third, when using the stars and bars approach, would this be a correct example for this problem:
.|...|..|..
First of all, when considering which of the numbers in the set $\{1, 2, \cdots 999999999\}$ have sum of digits equal to 13, you can pretend that the smaller numbers will be zero filled on the left, since including left-side zero digits does not affect the sum of the digits.
Second of all, it is impossible to have more than one specific digit equal to 9, and still have the sum of the digits equal to 13. Therefore, you can assume WLOG, that the leftmost digit is 9, obtain a computation, and then multiply this computation by 9 to get the final answer. The reason that you are applying the factor of 9 is that the 9-digit can occur in any of the 9 slots.
Using Stars and Bars analysis, as discussed here you want the number of non-negative solutions to
$$x_1 + x_2 + \cdots + x_8 = 4.$$
This computation is
$$\binom{4 + [8-1]}{8-1}.$$
Therefore, the final answer is
$$ 9 \times \binom{4 + [8-1]}{8-1} ~=~ 2970.$$