Combinatorics : Distributing p identical objects of one kind and q identical objects of other kind. In how many ways 5 blue pens and 6 black pens can be distributed to 6 children?

To do that I used:
$\text{Coefficient of } x^6 \text{ in } ((1+x+x^2+x^3+x^4+x^5)(1+x+x^2+x^3+x^4+x^5+x^6))$
and got answer = 6
but options given are:
a) 97020
b) 116424
c) 8008
d) 672
How does taking coefficient gives answer for distribution problems?
 A: It doesn’t. The coefficient of $x^6$ in your polynomial is the number of ways to pick $6$ of the $11$ pens. For instance, the term $x^2\cdot x^4$ (before you collect terms) corresponds to choosing $2$ blue and $4$ black pens.
HINT: This problem is quite different. Let’s forget about the black pens for a moment and ask in how many ways $5$ blue pens can be distributed amongst $6$ children. This is a standard stars and bars problem, and the answer is
$$\binom{5+6-1}{6-1}=\binom{10}5=252\;.$$
(The reasoning behind the formula is explained fairly well at the linked article.)
In similar fashion you can calculate the number of ways to distribute the $6$ black pens amongst the $6$ children. These two distributions are completely independent of each other: any distribution of the blue pens can be combined with any distribution of the black pens. This means that in order to get the total number of possible distributions of the $11$ pens, you should combine the number of distributions of the blue pens and the number of distributions of the black pens . . . how?
Note that all of this is based on the assumption that pens of the same color are indistinguishable, while the children are distinguishable. Thus giving George all $11$ pens is different from giving Tripta all $11$ pens, but giving Ivan $3$ blue pens and all the other pens to Nina is one arrangement no matter which $3$ blue pens Ivan gets.
A: If you denote by $B^n_m$ the number of ways to distribute n pens among m children, you can write down its generating series:
$$
\sum_{n=0}^\infty{B_m^nx^n}=(1+x+x^2+\ldots)^m=(1-x)^{-m}
$$
Now by taking the coefficient at $x^n$ you find a formula for $B_m^n$: 
$$
B_m^n=(-1)^n C_{-m}^n=C_{m+n-1}^n
$$
The answer to the problem is $B^5_6 B^6_6 = C_{10}^5 C_{11}^5 = 116424$.
A: Instead find the coefficient of $x^6 y^5$ in $(1+x+x^2+x^3+...x^\infty)^6(1+y+y^2+y^3+....y^\infty)^5$
I'll explain to you why taking the coefficient gives the answer for distribution problems. 
Consider the expression $(1+x+x^2+x^3+...x^\infty)^n$. You can write this expression as follows:$$(1+x+x^2+x^3+...x^\infty)(1+x+x^2+x^3+...x^\infty)(1+x+x^2+x^3+...x^\infty)...n \ \mathbb{times}$$
try to expand this expression. If $b$ denotes the coefficient of $x^m$ in this expression, then $$bx^m = \sum_{(i_1, i_2, i_3, ... i_n)}^{i_1+i_2+...+i_n = m}x^{i_1}x^{i_2}...x^{i_n}$$ where $i_k \in \mathbb{N} \cup \{ 0 \} $ for all $k \in \mathbb{N}$.
If you reflect on this for a while, you'll find that the coefficient $b$ will give you the number of ways in which $m$ can be sequentially partitioned into non - negative integers. And hence this coefficient will give you the number of ways in which $m$ objects can be distributed among $n$ distinct sets. 
P.S. : By sequentially partitioning, you can interpret that $(i_1, i_2, ... i_n)$ is different from $(i_2, i_1, ...i_n)$
