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Find the number of integer solutions to the equation:

$x(1) + x(2) + ··· + x(11) = 49 $

such that $0 \leqslant x(r) \leqslant 8$ for each $r= 1, 2,...,8$

I Used Principle Of Inclusion And Exclusion To get the following:

When any $1$ of the $x(r)$ are $ \geqslant 9$

When any $2$ of the $x(r)$ are $\geqslant 9$ ...

Until we are at-

When any $6$ of the $x(r) \geqslant 9$

And After This We Can't Proceed because the RHS would become negative ,

So I Am not sure what to do next...

Edit: I Want the Solution From The Ball And Box Principle And P.I.E. Contrary as Mentioned Below In The Comments

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  • $\begingroup$ HINT Firstly, you need to apply stars and bars, and then apply inclusion- exclusion, starting with $\displaystyle{\binom{49+11-1}{11-1}}\;$ thus $\displaystyle{\binom{59}{10} - \binom{11}1\binom{50}{10} +...}$ $\endgroup$
    – true blue anil
    Commented Mar 21, 2023 at 4:09
  • $\begingroup$ The generic form of your question has been asked over and over again, on MathSE. See this answer for a blueprint of how to combine Inclusion-Exclusion with Stars-And-Bars to attack this generic type of problem. $\endgroup$
    – user2661923
    Commented Mar 21, 2023 at 6:45
  • $\begingroup$ Re my last comment, I am therefore voting to close this question, as a duplicate. $\endgroup$
    – user2661923
    Commented Mar 21, 2023 at 6:46

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