How many ways $5$ identical green balls and $6$ identical red balls can be arranged into $3$ distinct boxes such that no box is empty? 
How many ways $5$ identical green balls and $6$ identical red balls can be arranged into $3$ distinct boxes such  that no box is empty?

My attempt :
Finding coefficient of $x^{11}$ in the expansion of $$( x + x^2 + x^3 + x^4 + x^5+x^6 )^3 ( x + x^2 + x^3 + x^4 + x^5 )^ 3$$  and arranging them which was wrong when inspected
Please help me out
 A: Strategy:
First distribute the green balls, then consider cases depending on how many boxes are left empty.
Two boxes are left empty:  This occurs if all the green balls are placed in one box.  There are three ways this can occur.  Place a red ball in each of the other two boxes so that no box is left empty.  Then distribute the remaining four red balls to the three boxes without restriction, which reduces to solving the equation
$$x_1 + x_2 + x_3 = 4$$
in the nonnegative integers, where $x_i$ is the number of red balls placed in the $i$th box, $1 \leq i \leq 3$.
Since a particular solution of  the equation
$$x_1 + x_2 + \cdots + x_k = n$$
in the nonnegative integers corresponds to the placement of $k - 1$ addition signs in a row of $n$ ones, the number of solutions of the equation in the nonnegative integers is
$$\binom{n + k - 1}{k - 1}$$
since we must choose which $k - 1$ of the $n + k - 1$ positions required for $n$ ones and $k - 1$ addition signs will be filled with addition signs.  See Theorem 2.
One box is left empty: This occurs if four green balls are placed in one box and one green ball is placed in another or three green balls are placed in one box and two green balls are placed in another.  Each of these distributions can occur in $3 \cdot 2 = 6$ ways.  Place a red ball in the empty box so that no box is left empty.  Then distribute the remaining five red balls to the three boxes without restriction, which reduces to solving the equation
$$x_1 + x_2 + x_3 = 5$$
in the nonnegative integers, where $x_i$ is the number of red balls placed in the $i$th box, $1 \leq i \leq 3$.
No box is left empty:  The number of ways the five green balls can be distributed to the three boxes so that no box is left empty is the number of solutions of the equation
$$x_1 + x_2 + x_3 = 5$$
in the positive integers, where $x_i$ is the number of balls in the $i$th box, $1 \leq i \leq 3$.
A particular solution of the equation
$$x_1 + x_2 + \cdots + x_k = n$$
in the positive integers corresponds to the placement of $k - 1$ addition signs in the $n - 1$ spaces between successive ones in a row of $n$ ones.  The number of such solutions is
$$\binom{n - 1}{k - 1}$$
since we must choose which $k - 1$ of those $n - 1$ spaces will be filled with addition signs. See Theorem 1.
The six red balls can then be distributed to the three boxes without restriction, which reduces to solving the equation
$$x_1 + x_2 + x_3 = 6$$
in the nonnegative integers, where $x_i$ is the number of red balls placed in the $i$th box, $1 \leq i \leq 3$.  Apply Theorem 2 to do this.
A: When we neglect the nonempty condition we can distribute the $5$ green balls in ${5+2\choose2}=21$ ways into the three distinct boxes, and independently of this the $6$ red balls in ${6+2\choose2}=28$ ways. This gives $21\cdot28=588$ possible distributions.
In the same way we can compute the number of distributions where the third box has to remain empty. This gives a total of ${5+1\choose1}\cdot{6+1\choose1}=42$ distributions, and the same number arises when another box has to remain empty.
The total number $N$ of admissible distributions therefore is given by
$$N=588-3\cdot 42+3=465\ .$$
At the end we have added $3$, because the three distributions where two boxes remain empty have been subtracted twice in the $3\cdot42$ term.
