Find all $4$ digit numbers such that sum of digits is $11$. \begin{cases} x_1+x_2+x_3+x_4=11\\ 1\leq x_1{\leq 9}\\ 0\leq x_2{\leq 9}\\0\leq x_3{\leq 9}\\ 0\leq x_4{\leq 9}\end{cases}.
Using stars and bars there is 13 choose 3 ways to do. So we get $286$ but answer is $279$. I saw this same question in the forum but didn't fully understand solution. Where we are overcounting when solving in this way? And mainly what we are overcounting?
As stated by Stinking Bishop in the comments, you are overcounting e.g. the "number" $(x_1, x_2, x_3, x_4) = (11, 0, 0, 0)$. The problem is that your stars and bars method does not account for your stated constraints $0\leq x_i\leq 9$.
Therefore, we need to subtract the extra cases where one of the digit is at least $10$. If there's a $10$, it must be some permutation of $(10, 1, 0, 0)$, so there are $6$ cases (note that $0$ cannot be first digit). If there's a $11$, then it must be $(11, 0, 0, 0)$, so there is $1$ case here. Therefore, there is $286 - 6 - 1 = 279$ cases in total.
Challenge for you: Do the same problem for sum of digits $S = 12$ and $13$. There should be $342$ and $405$ respectively.
In this answer it is described how to solve this by generating functions. The solution is the coefficient of the degree $11$ monomial of the polynomial
$$(x+x^2+\cdots+x^9)(1+x+x^2+x^3+\cdots+x^9)^3$$
This coefficient can be calculated by a CAS like Mathematica as shown in the link above. It can even calculated by a numerical calculator that can handle arbitrary length integers as describe in the second half of my answer here. But in this answer I learned in one of the comments how to calculate this by hand.
Note that
$$\frac 1 {1-x}=\sum_{n=0}^\infty x^n$$ and by differentiation using induction you get
$$\frac 1 {(1-x)^k}=\sum_{n=0}^\infty{ n+k-1 \choose k-1} x^n$$
So we have
$$(x+x^2+\cdots+x^9)(1+x+x^2+x^3+\cdots+x^9)^3\\ =x\frac{1-x^9}{1-x}\frac{(1-x^{10})^3}{(1-x)^3}\\ =x(1-x^9)(1-x^{10})^3 \sum_{n=0}^\infty{ n+3 \choose 3} x^n\\ =(x-x^{10}-3x^{11}+3x^{20}+\cdots)\frac 1 6 (1\cdot2\cdot3+2\cdot 3 \cdot4\cdot x+\cdots+11\cdot 12\cdot 13\cdot x^{10} \cdots )\\ =\cdots +\cdot \frac 1 6 (-3\cdot 1\cdot 2 \cdot 3-2\cdot 3\cdot 4+11\cdot 12 \cdot 13)x^{11}+\cdots\\ =\cdots+279x^{11}+\cdots$$
The coefficients of the monomials of degree greater than $36$ cancel out.
We have
$$ \frac{\left( {{x}^{40}}-{{x}^{31}}-3 {{x}^{30}}+3 {{x}^{21}}+3 {{x}^{20}}-3 {{x}^{11}}-{{x}^{10}}+x\right) \, \sum_{n=0}^{\infty }{\left. \left( {{n}^{3}}+6 {{n}^{2}}+11 n+6\right) {{x}^{n}}\right.}}{6} $$
We abbreviate $$f(n):={{n}^{3}}+6 {{n}^{2}}+11 n+6$$ and verify that $$f(n+39)-f(n+30)-3f(n+29)+3f(n+20)+3f(n+19)-3f(n+10)-f(n+9)+f(n)=0$$ and $$f(n+39)-f(n+30)-3 f(n+29)+3 f(n+20)+3 f(n+19)-3 f(n+10)-f(n+9)$$ for $n=-1,-2,-3$ because $f(n)=0$ for this values
(x*(1-x^9)*(1-x^10)^3*sum(binomial(n+3,3)*x^n,n,0,oo)).full_simplify() gives a polynomial of degree $36$. Sage is great :D
– Gareth Ma
May 25 '22 at 14:09
As is usual, in using stars and bars, we often need to apply inclusion-exclusion if there is an upper limit.
Here there are also differing lower limits, so we will first equalize the lower limits by preplacing a digit in $x_1$, so
$\;x_1+x_2+x_3+x_4 = 10,$
and upper limit becomes $8$ for $x_1$ and $9$ for the rest, so if we put $9$ in the first digit or $10$ in any other, they are the "bad" cases to be excluded.
and we get $\binom{13}3 -\binom 11\binom{13-9}3 - \binom31\binom{13-10}3 = 279$
The other alternative is to use generating functions, and find the coefficient of $x^{11}$ in $(x^1+x^2+x^3+...+x^9)(x^0+x^1+x^2+x^3+...+x^9)^3$
with the first multiplicand standing for the first digit, yielding the same answer of $279$
