What is the number of possibilities to choose 80 numbers out of the set $~\{10,11,\cdots,99\}~$ with repetition and no order significant. In which if an element that divides by $10$ with no Remain of division selected it must be chosen at least as many times as its $~10^{\text{th}}~$ digit.
Example: if $~30~$ is chosen it will be chosen minimum $~3~$ times. Otherwise it could not be selected.
I've been tried to solve it with Inclusion-Exclusion by taking the sum of all the possibilities and reduce from it the following groups:
$A_2=$ all the possibilities that $20$ will appears only once $\binom{89}{79}$
$A_3=$ all the possibilities that $30$ will appears only once, or only twice $\binom{89}{79}$ + $\binom{89}{78}$.
.
.
.
$A_9$
which give me the following :
$$\binom{89}{79}-\Biggl(\sum_0^1 \binom{89}{79-i} + \sum_0^2 \binom{89}{79-i} + \sum_0^3 \binom{89}{79-i} +\sum_0^4 \binom{89}{79-i}+\cdots+\sum_0^8 \binom{89}{79-i}\Biggr)$$ and so on..
I think, I'm missing the target by trying way too hard to manipulate the problem into this formula and it sure has a simpler way to calculate it. but I can't put my finger on it.
Please advice.
You are asking for the number of non-negative integer solutions of $x_{10}+x_{11}+\dots+x_{99}=80$ subject to $x_i\geq 0, x_{10}\geq 1, x_{20}\geq 2,\dots,x_{90}\geq 9$.
Consider a change of variable., letting $y_{10}=x_{10}-1$, letting $y_{20}=x_{20}-2$, ..., $y_{90}=x_{90}-9$ and $y_i=x_i$ otherwise.
You are now looking for the number of non-negative integer solutions of $y_{10}+y_{11}+\dots+y_{99}=35$ subject to $y_i\geq 0$ for each $i$.
This can be solved using traditional stars-and-bars approaches.
If this question were reflavored as asking how many ways you can distribute cookies to children, where the tenth child must get at least one cookie, the twentieth child must get at least two cookies, etc... you can explain the solution above by "giving the required necessary cookies out ahead of time and then distributing the remaining cookies after the fact."