Sum of Coefficients and Number of Terms in Trinomials and Quadrinomials I already know how to find the sum of coefficients in a binomial, but how do you do it for a trinomial/quadrinomial (after like terms are added)?
Example Problem: $(wa+xb+yc+zd)^n$ (all variables are integers)
I have the same issue for the number of terms in trinomials and quadrinomials (after like terms are added). I was supposed to do the problem using the example above as well. 
 A: The idea is essentially the same as in the binomial case: To expand $$(X_1 + \cdots + X_m)^n,$$
one determines the possible products of the $X_i$ that can occur, namely all of the terms
$$X_1^{k_1} \cdots X_m^{k_m}$$
with total degree $k_1 + \cdots + k_m = n$ in the monomials, and the Multinomial Theorem tells you that the coefficients of those terms are respectively
$${{n}\choose{k_1, \ldots, k_m}} = \frac{n!}{k_1! \cdots k_n!},$$
which yields the expansion
$$(X_1 + \cdots + X_m)^n = \sum_{k_1 + \cdots + k_m = n} \frac{n!}{k_1! \cdots k_n!} X_1^{k_1} \cdots X_m^{k_m},$$
where the sigma notation here just indicates that we add the terms ${{n}\choose{k_1, \ldots, k_m}} X_1^{k_1} \cdots X_m^{k_m}$ for each list $(k_1, \ldots, k_m)$ of $m$ nonnegative integers whose sum is $n$.
Example The simplest nontrivial case not covered by the binomial theorem is the square of a trinomial:
$$(X + Y + Z)^2.$$
The terms that can occur here are all of the quadratic monomials in $X, Y, Z$, namely,
$$X^2, Y^2, Z^2, YZ, ZX, XY,$$
and we can easily compute their respective coefficients:
For emphasis, we can write $X^2 = X^2 Y^0 Z^0$, so the coefficient of $X^2$ is
$${{2!}\choose{2!0!0!}} = \frac{2!}{2!0!0!} = \frac{2}{2} = 1.$$
By symmetry, this is also the coefficient of the $Y^2$ and $Z^2$ terms.
Similarly, we can write $YZ = X^0 Y^1 Z^1$, so the coefficient of $YZ$ is
$${{2!}\choose{0!1!1!}} = \frac{2!}{0!1!1!} = \frac{2}{1} = 2,$$
and again, this is also the coefficient of the $ZX$ and $XY$ terms.
This accounts for all of the terms, so we get
$$(X + Y + Z)^2 = X^2 + Y^2 + Z^2 + 2YZ + 2ZX + 2XY.$$
Remark In practice, often when one uses the Multinomial Theorem, one doesn't need the full expansion of a power of a multinomial but rather just a coefficient of a single term.
A: I'm not allowed to comment just yet so I'm putting this up as answer (to the first part of your question only).
To find the sum of coefficients in a polynomial expansion, simply add each term's in the original expression and raise the sum to the expression's exponent.
For instance, you want to find the sum of coefficients in the expansion $(4x+3y-2z)^8$.The answer would be $(4+3-2)^8=390,625$.
