Number of terms$(2x+3y-4z)^n$ What is the number of terms in the expansion of $(2x+3y-4z)^n$?
For terms of type $(x+y)^n$ we have $n+1$ terms but for three type of term I am not able to solve it. I know the answer but not able to derive it
 A: The number of terms in $(x+y)^n$, when expanded is $n+1$.  Now, if we look at $(2x+3y-4z)^n$ and let $w=2x+3y$, then you have $(w-4z)^n$, which has $n+1$ terms.
Each of these terms is of the form $C(2x+3y)^a(-4z)^{n-a}$, where $C$ is an appropriate constant.  Since each of these terms has a different power on $z$, when you expand the first part, there will be no cancellation.
Since $a$ can vary between $0$ and $n$, and we know how many terms $(2x+3y)^a$ expands into, we get that the number of terms is
$$
1+2+\dots+(n+1)=\sum_{a=0}^n a+1.
$$
This has a standard formula and is
$$
\frac{(n+1)(n+2)}{2}.
$$
A: A general term in the expansion has the form 
$$2^{k_1}x^{k_1}\,3^{k_2}y^{k_2}\,(-4)^{k_3}z^{k_3},\quad \text{satisfying the condition}\quad k_1+k_2+k_3=n,  $$
so the number of terms is just the number of solutions of the equation in natural numbers: 
$k_1+k_2+k_3=n$. This is a special case of the general equation:
$$k_1+k_2+\dots+k_r=n,\qquad k_1,k_2,\dots,k_r\in\mathbf N. $$
It can be shown by induction the number of solutions of the general equation is equal to
$$S(n,r)=\binom{n+r-1}{r-1}.$$
