How many collected terms are in the expansion of $(x+y+z)^{10} (w+x+y+z)^2$?
Hi, I'm trying to solve this problem as study material for discrete mathematics and I'm not quite sure how. I got 235 terms by plugging it into Wolfram Alpha, but I'm don't really know how to get the answer. I feel like combinatorics and the binomial theorem can be applied in some way as per the curriculum in my class, but that's about it.
First of all, observe that:
$$(x+y+z)^{10}(w+x+y+z)^2=w^2(x+y+z)^{10}+2w(x+y+z)^{11}+(x+y+z)^{12};$$
and notice that the total number of terms of $(x+y+z)^{10}(w+x+y+z)^2$ is the sum of those of $w^2(x+y+z)^{10}, 2w(x+y+z)^{11}$, and $(x+y+z)^{12}$ because of the distinct powers of $w$.
Hence the answer is $\begin{pmatrix} 12 \\ 2 \\ \end{pmatrix}+\begin{pmatrix} 13 \\ 2 \\ \end{pmatrix}+\begin{pmatrix} 14 \\ 2 \\ \end{pmatrix}=235.$ For more information you can read this link and this one.