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Here's what I have done: since 5 digits are at least 1, that leaves the value 17-5 = 12 to be distributed in the rest of the digits so right now, the number is 11,111. the value 12 is distributed into the 5 digits, and using the stars and bars method, the answer should be 16C4 = 1820

What am I doing wrong (btw, the answer shows as 1645, but I don't know how)

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    $\begingroup$ You can't use Stars and Bars directly, since the digits are capped. You'll have to exclude those combinations which would require a digit $>9$. $\endgroup$
    – lulu
    Dec 30, 2023 at 14:49
  • $\begingroup$ To verify that 1645 is correct, with brute force, you can use PARI/GP in which false is represented as 0, and true as 1, and therefore the expression sum(n=10000,99999,d=digits(n);vecmin(d)>0&&vecsum(d)==17) gives you the result (indeed 1645). Of course, this is not an answer. $\endgroup$ Dec 31, 2023 at 12:08
  • $\begingroup$ Extended problem: Find the number of good numbers regardless of digit count (number of digits could be anything between 2 and 17). $\endgroup$ Dec 31, 2023 at 12:14

6 Answers 6

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You have to exclude the numbers with “digits” equal to 10, 11, 12, 13 from consideration. Those numbers are (with random digits order):

$13, 1, 1, 1, 1$

$12, 2, 1, 1, 1$

$11, 3, 1, 1, 1$

$11, 2, 2, 1, 1$

$10, 4, 1, 1, 1$

$10, 3, 2, 1, 1$

$10, 2, 2, 2, 1$

The quantities of those numbers are the following:

$5$

$5 \cdot 4 = 20$

$5 \cdot 4 = 20$

$5 \cdot 6 = 30$

$5 \cdot 4 = 20$

$5 \cdot 4 \cdot 3 = 60$

$5 \cdot 4 = 20$

Substracting these numbers from your answer $1820$ gives the right answer $1645$

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We can solve this question using fairly simple combinatorics.

Let each digit be represented as $A, B, C, D$ and $E$. In this manner, 1 $\leq A \leq 9$, 1 $\leq B \leq 9$, and so on.

This is the equivalent of finding the coefficient of $x^{17}$ in

$$(x^{1}+x^{2}+x^{3}+x^{4}+x^{5}+x^{6}+x^{7}+x^{8}+x^{9})^{5}$$ We can further simplify this as:

$$x^{5}(1-x^{9})^{5}(1-x)^{-5}$$

So, coefficient of $x^{17}$ becomes $$^{5+12-1}C_{12}\times ^{5}C_{0} - ^{5}C_{1} \times ^{5+3-1}C_{3}$$

$$=^{16}C_{12}-5\times ^{7}C_{3}$$

$$=1645$$

Hope this helped.

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This is a kind of fun answer. Let M be the matrix $$\left( \begin{array}{ccccccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \end{array} \right).$$ The answer for finding the number of $k$ digit numbers whose digits do not include 0 and sum to $k+j$ is equal to the first element in the $j$th row of $M^k$ if $j<=12$.

Polynomial Method Another very similar method is to use powers of $f(z) = z + z^2 +\cdots + z^9$.
The number of $k$ digit numbers whose digits do not include 0 and sum to $j$ is equal to the coefficient of $z^j$ of $(f(z))^k$. Note that $$f(z) =\frac{z^{10}-1}{z-1}-1$$ and that for any polynomial $p(z)$, the coefficient of $z^j$ is $$c = p^{(j)}(0)/j!$$ where $p^{(j)}$ is the $j$th derivative of $p$.

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As indicated by the comment of lulu, if Stars and Bars is to be used, it must be combined with Inclusion-Exclusion. I will follow the approach taken in this answer.

You want the number of solutions to

  • $~x_1 + x_2 + x_3 + x_4 + x_5 = 17.$

  • $~x_1, ~x_2, ~x_3, ~x_4, ~x_5 \in \Bbb{Z_{\geq 1}}.$

  • $~x_1, ~x_2, ~x_3, ~x_4, ~x_5 \in \Bbb{Z_{\leq 9}}.$

Stars and Bars theory is based on the lower bound being $~0,~$ rather than $~1.$ So, first, use the change of variable
$~y_i = x_i - 1 ~: ~i \in \{1,2,3,4,5\}.~$

So, now you want the number of solutions to

  • $~y_1 + y_2 + y_3 + y_4 + y_5 = 17 - 5 = 12.$

  • $~y_1, ~y_2, ~y_3, ~y_4, ~y_5 \in \Bbb{Z_{\geq 0}}.$

  • $~y_1, ~y_2, ~y_3, ~y_4, ~y_5 \in \Bbb{Z_{\leq 8}}.$

Let $~S~$ denote the set of all solutions, where the third bullet point above is ignored.

For $~i \in \{1,2,3,4,5\}, ~$ let $~S_i~$ denote the subset of $~S~$ where the specific variable $~y_i~$ is in violation (i.e. $~y_i \geq 9~$) and the the other variables may or may not be in violation.

Then, the desired computation is

$$|S| - |S_1 \cup S_2 \cdots \cup S_5|.$$

Notice that in this case, it is impossible for more than one variable to be in violation. That is, given any distinct $~i,j \in \{1,2,3,4,5\},~$ you must have that $~S_i \cap S_j = \emptyset.~$

This is because $~(2 \times 9) > 12.$

So, by Inclusion-Exclusion theory, and by considerations of symmetry,

$$|S| - |S_1 \cup S_2 \cdots \cup S_5| = |S| - 5|S_1|.$$

To compute $~|S_1|,~$ you make the further change of variable $~z_1 = y_1 - 9.~$

Therefore, $~|S_1|~$ equals the number of solutions to

  • $~z_1 + y_2 + y_3 + y_4 + y_5 = 12 - 9 = 3.$

  • $~z_1, ~y_2, ~y_3, ~y_4, ~y_5 \in \Bbb{Z_{\geq 0}}.$

So, By Stars and Bars Theory,

$$|S| - 5|S_1| = \binom{16}{4} - 5\binom{7}{4} = 1645.$$

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Proof by brute force and dynamic programming:

[1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5, 4, 3, 2]
[0, 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 52, 57, 60, 61, 60, 57]
[0, 0, 0, 1, 4, 10, 20, 35, 56, 84, 120, 165, 216, 270, 324, 375, 420]
[0, 0, 0, 0, 1, 5, 15, 35, 70, 126, 210, 330, 495, 710, 976, 1290, 1645]

The $k$-th term on line $n$ is the number $X_k^n$ of ways to obtain sum $k$ with $n$ digits in $1,\ldots,9$. The induction formula is \begin{equation} X_k^{n+1} = \sum_{i=1}^9 X_{k-i}^n \end{equation} The table shows that $X_{17}^5 = 1645$

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@Amritraj Lamba: Hi welcome to MSE. You can turn the problem into an equation like below $$ \overline{abcde} \to a+b+c+d+e =17$$ but some conditions are missed here, what conditions? $$ 1\leq a,b,c,d,e \leq 9$$ then ude combinatorics to find the numbers of solutions. $$(a-1)+(b-1)+(c-1)+(d-1)+(e-1)=17-5 \\a'+b'+c'+d'+e'=12\\ {{14+5-1}\choose {5-1}}-5{2+5-1 \choose 5-1}$$

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    $\begingroup$ The conditions are $1\le a, b, c, d, e\le 9$ $\endgroup$ Dec 30, 2023 at 15:07
  • $\begingroup$ Oh thanks, I fixed it. $\endgroup$
    – Khosrotash
    Dec 30, 2023 at 15:17
  • $\begingroup$ Python gives $\binom{18}{4} - 5 \binom{6}{4}=2985$ which is not the correct result $1645$. $\endgroup$ Dec 30, 2023 at 16:02

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