Binomial coefficient complex expression I have been trying to find the coefficient of
$a^2x^3$, in the expansion of $(a+x+c)^2(a+x+d)^2$, without success. 
I am having trouble expanding the above expression, because I can't find a way to merge them into one.
How can I solve this? Thanks
 A: Well, I would just do this by `brute force', as they say, with a lot of distributing:
$(a+x+c)^2(a+x+d)^2=(a+x+c)(a+x+c)(a+x+d)(a+x+d)=(a+x+c)(a+x+c)[a(a+x+d)+x(a+x+d)+d(a+x+d)]$
and keep going from there. Does this pattern make sense?
A: Here is a variation to determine the coefficient of $a^2x^3$ without a full expansion of the expression. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in an expression.

We obtain
  \begin{align*}
[a^2x^3]&(a+x+c)^2(a+x+d)^2\\
&=[a^2]\left([x^2](a+x+c)^2\right)\left([x^1](a+x+d)^2\right)\\
&\qquad +[a^2]\left([x^1](a+x+c)^2\right)\left([x^2](a+x+d)^2\right)\tag{1}\\
&=[a^2]\left(2(a+d)\right)+[a^2]\left(2(a+c)\right)\tag{2}\\
&=0
\end{align*}

Comment:


*

*In (1) we observe that in order to obtain the coefficient of $x^3$ one factor has to contribute $x$ and the other factor has to contribute $x^2$. There are no other possibilities to obtain a term with $x^3$.

*In (2) we select the coefficient of $x$ resp. $x^2$ according to the binomial formula
\begin{align*}
(a+x+u)^2=(a+u)^2+2(a+u)x+x^2
\end{align*}
Since there is no term with a factor $a^2$ the result is $0$.
A: You have $(a + x + c)^2(a + x + d)^2$ and want to find the coefficient of $a^2x^3$. In other words, you have four boxes: the first two each have 2 $a$s, 2 $x$s, and 2 $c$s, while the second two each have 2 $a$s, 2 $x$s, and 2 $d$s. You want to pick exactly one letter from each box and in the end have 2 $a$s and 3 $x$s. Obviously, this isn't possible because you need 5 letters, and thus the coefficient is 0.
(Suppose that you instead wanted the coefficient of $a^2x^2$. Then you would choose 2 of the 4 boxes to pick an $a$ from, and pick a $d$ from the other 2. This means the coefficient is the number of ways you can choose 2 from 4, $\binom{4}{2} = 6$.
What about $ac^2d$? The 2 $c$s can only come from the first two boxes, so you can take an $a$ and a $d$ from the second two, in a total of 2 ways.
This method is generalizable.)
A: It is easy to see that all monomials in you expression are of form $a^{\alpha}c^{\gamma}d^{\delta}x^{\chi}$ for some non-negative integers $\alpha$, $\gamma$, $\delta$ and $\chi$ such that $\alpha + \gamma + \delta + \chi = 4$. You need coefficient of $a^2x^3$, where sum of powers is not 4, therefore coefficient is 0.
