Distributing $10$ indistinguishable cars, $12$ indistinguishable balls, $14$ indistinguishable teddy bears to $3$ children, each have exactly $7$ toys 
I have $10$ indistinguishable cars , $12$ indistinguishable balls, $14$ indistinguishable teddy bears.I want to distribute them to $3$ different children in a kindergarten such that each child will take exactly $7$ toys.How many distributions are there?

$\mathbf{\text{My attempt:}}$ First of all, I thought that it can be solved by generating functions easily. However, the process became suddenly cumbersome.
If each child will take exactly $7$ toys , then we need $21$ toys in total.  For example , $10$ cars, $6$ balls and $5$ teddy bears can be a sample. By using this information, I decided to use generating functions such that if a child take exactly $7$ toys, then his generating function form is $$(x^7+y^7+z^7+x^6y+x^6z+x^5y^2+x^5yz+...+yz^6)$$ where $x,y,z$ represents cars, balls and teddy bears, respectively.  As you count by combination with repetition, there are $\binom{7+3-1}{7}=36$ terms in the tuple.
Because of there are $3$ children , we will deal with $$(x^7+y^7+z^7+x^6y+x^6z+x^5y^2+x^5yz+...+yz^6)^3$$
However, there is a problem for me. Calculating the coefficients for each possible selection is very cumbersome. For example, we said that one of the possible toy selections of $21$ of $36$ is $10$ cars, $6$ balls and $5$ teddy bears. In that sample, we should find $$[x^{10}y^6z^5](x^7+y^7+z^7+x^6y+x^6z+x^5y^2+x^5yz+...+yz^6)^3$$
As you realize there are many other toy selections out of $36$, for example $5$ cars, $8$ balls and $8$ teddy bears is an another sample.
I want you guys to handle this cumbersome selection process.  How can I find the all coefficients of this generating function for all possibilities?
Please do not suggest using a computer algorithm. I am here to see a mathematical approach I could apply to my problem. What's more, if you have another approach, feel free and share it with me. I am open to another methods.
 A: Method 1
Instead of using variables for the toys, use variables for the kids. This is the right idea since there is only way to satisfy the kids, $(7,7,7)$, while there are many ways to satisfy the toys.
Let
$$
F_{\text{car}}(x,y,z)=\sum_{\substack{i,j,k \,\ge \,0,\\ i+j+k\,\le\, 10}} x^iy^jz^k
$$
be the generating function which describes all ways to distribute the cars to the kids. Define $F_{\text{ball}}$ and $F_{\text{teddy}}$ similarly, where $10$ is replaced with $12$ and $14$. Then
$$
\text{# ways}=[x^7y^7z^7](F_{\text{car}}\times F_{\text{ball}}\times F_{\text{teddy}})
$$
Method 2
Here is an equivalent version of your problem. You have $10+12+14=36$ identical tokens, and $12$ urns. The urns are laid out in a $4\times 3$ grid. Three of the rows correspond to children, while the last row is a slack row that represents the leftover toys no child receives. Each column corresponds to a toy. The equivalent problem is:

How many ways are there to distribute the $36$ identical tokens to these $12$ urns so that there are $7$ tokens total in each of the "child" rows, and $10$ (resp. $12,14$) tokens total in the "car" (resp. "ball," "teddy") columns?

Since there are now six target totals to keep track of, we need six variables. I will use $x,y,z$ for the children again, and $a,b,c$ for the teddy, ball, and car respectively.  Since there are $12$ urns, we need twelve generating functions. The generating functions are summarized in the below table.
The final answer is the coefficient of $x^7y^7z^7a^{14}b^{12}c^{10}$ in the product of all twelve functions in this table.





Teddy $(a)$
Ball $(b)$
Car $(c)$




Child 1 $(x)$
$1/(1-ax)$
$1/(1-bx)$
$1/(1-cx)$


Child 2 $(y)$
$1/(1-ay)$
$1/(1-by)$
$1/(1-cy)$


Child 3 $(z)$
$1/(1-az)$
$1/(1-bz)$
$1/(1-cz)$


Unused
$1/(1-a)$
$1/(1-b)$
$1/(1-c)$




For example, the generating function for urn for child $1$ and teddy is $$1/(1-ax)=1+ax+a^2x^2+a^3x^3+\dots$$ Choosing a monomial from this series is equivalent to choosing the number of tokens that go into the urn in the child $1$ row and teddy column. If you put $n$ tokens in that urn, it is equivalent to choosing $a^nx^n$, which contributes $n$ to the quota of $x^7$, and contributes $n$ to the quota of $a^{14}$.
Computation
As a sanity check, we should see if these methods agree. The following Mathematica code indeed confirms a common answer of $35{,}336$ for both methods. The second method is much faster; the first took $30$ seconds on my machine, the latter only half a second.
(* Method 1 *)

F[x_, y_, z_, n_] := Sum[x^i y^j z^k Boole[i + j + k <= n], 
                         {i, 0, n}, {j, 0, n}, {k, 0, n}]

Print[Coefficient[F[x, y, z, 10] * F[x, y, z, 12] * F[x, y, z, 14], x^7 y^7 z^7]]

(* Method 2 *)

Print[SeriesCoefficient[
  ((1 - a x) (1 - b x) (1 - c x)
   (1 - a y) (1 - b y) (1 - c y)
   (1 - a z) (1 - b z) (1 - c z)
   (1 - a)   (1 - b)   (1 - c)  )^{-1},
  {x, 0, 7}, {y, 0, 7}, {z, 0, 7}, {a, 0, 14}, {b, 0, 12}, {c, 0, 10}
]]

