The number of ways can $10$ balls be chosen from an urn containing $10$ identical green balls, $5$ identical yellow balls and $3$ identical blue balls.
Can someone guide me how should I solve such problems?
I tried with combination with repetition but I am getting confused with it.
Let $x_g, x_y, x_b$ represent the number of green balls chosen, yellow balls chosen, and blue balls chosen respectively.
We have then the system:
$$\begin{cases} x_g+x_y+x_b=10\\ 0\leq x_g\leq 10\\ 0\leq x_y\leq 5\\ 0\leq x_b\leq 3\end{cases}$$
We wish to count the number of integer solutions to this system.
Let $A_g,A_y,A_b$ each represent the set of solutions which violate the upper bound condition on green, yellow, and blue balls selected respectively (among possibly violating additional upper bound conditions as well). Let $S$ represent the set of solutions where we don't care about the upperbound conditions at all.
We are tasked then with calculating $|A_g^c\cap A_y^c\cap A_b^c|$ which by inclusion-exclusion is equal to:
$$|S|-|A_g|-|A_y|-|A_b|+|A_g\cap A_y|+|A_g\cap A_b|+|A_y\cap A_b|-|A_g\cap A_y\cap A_b|$$
To continue, remember that the number of integer solutions to the system:
$$\begin{cases} x_1+x_2+\dots+x_r = n\\ 0\leq x_i~~~\forall i\end{cases}$$
is equal to $\binom{n+r-1}{r-1}$
and note that each of the above can be transformed into a calculation of this form via a change of variable and shifting of indices.
For example, calculating $|A_y\cap A_b|$ we count the number of solutions which violate both the upperbound on yellow and blue simultaneously, so we have the system:
$$\begin{cases}x_g+x_y+x_b=10\\0\leq x_g\\ 6\leq x_y\\ 4\leq x_b\end{cases}$$
which is equivalent to the system:
$$\begin{cases}y_g+y_y+y_b=0\\0\leq y_g\\ 0\leq y_y\\ 0\leq y_b\end{cases}$$
The number of solutions to this specific case is $\binom{3-1}{3-1}=\binom{2}{2}=1$, corresponding to the one outcome where we have too many of both yellow and blue balls, i.e. the outcome (0 green, 6 yellow, 4 blue)
$\binom{12}{2}-0-\binom{6}{2}-\binom{8}{2}+0+0+\binom{2}{2}-0=24$