QUESTION: How many terms are there in the expansion of $$(x+y)(a+b+c)(e+f+g)(h+i)$$
I'd like some help with this one, but I'd also like to discuss a method of generalization on the problem, namely how should one compute the cardinality of terms for something like this:
$$\Sigma_{i=0}^{\alpha_1}\zeta_{1_i}\cdot\Sigma_{i=0}^{\alpha_2}\zeta_{2_i}\cdots\Sigma_{i=0}^{\alpha_n}\zeta_{1_n}$$
Here each $\alpha_k$ is arbitrary, so there are any number of $\zeta_{j_i}$'s for each sum.
$$\text{How does this sound?}$$ $$\downarrow$$
Let us generalize the problem a bit. We are asked to find the cardinality of terms in the expansion of a product of multinomials. This can be represented in the following way: $$\sum_{i=0}^{\alpha_0}\zeta_{0_i}\cdot\sum_{i=0}^{\alpha_1}\zeta_{1_i}\cdots \sum_{i=0}^{\alpha_n}\zeta_{n_i}$$ Now, consider the product of two such multinomials, say $$(\zeta_{0_0}+\zeta_{0_1}+\cdots+\zeta_{0_{\alpha_0}})(\zeta_{1_0}+\zeta_{1_1}+\cdots+\zeta_{1_{\alpha_1}}).$$ From this it is clear that the first term in the first multinomial---that is, $\zeta_{0_1}$--- will be multiplied by the $\alpha_1$ terms in the latter multinomial, and so there will thus far be $\alpha_1$ distinct terms generated from this process. Now, this process will be done on each of the $\zeta_{0_i}$'s exactly $\alpha_0$ times, so it is quite clear that the number of distinct terms generated from these processes is precisely $\alpha_0 \cdot\alpha_1$. By extension, this action could then be done to a third multinomial with $\alpha_2$ terms, and so by the above logic this would give $\alpha_0\cdot\alpha_1\cdot\alpha_2$ distinct terms in the final expansion of three such multinomials. From this we see that the cardinality of the terms in the expansion of a product of multinomials is $$\prod_{j=0}^{n}\alpha_j,$$ so if we are asked to find the cardinality of the terms in the expansion of $$(x+y)(a+b+c)(e+f+g)(h+i),$$ then we notice that from left to right we have that $\alpha_0=2$, $\alpha_1=3$, $\alpha_2=3$, and $\alpha_3=2$, which leads us to the conclusion that the desired cardinality is $\alpha_0\cdot\alpha_1\cdot\alpha_2\cdot\alpha_3=2\cdot 3\cdot 3\cdot 2=36$, namely there are exactly $36$ terms in the expansion of the expression above.
You can get everything you need by using the fact that $(x_1+x_2+\cdots +x_m)(y_1+y_2+\cdots +y_n)$ has $mn$ terms.
To show this, note that the terms are $x_1$ times stuff ($n$ terms), and $x_2$ times stuff ($n$ more terms, and so on, for a total of $\underbrace{n+n+\cdots +n}_{\text{$m$ times}}$; that is, $m$ copies of $n$.
Now you should be able to deal with $(x_1+\cdots+x_m)(y_1+\cdots +x_n)(z_1+\cdots+x_p)$. Multiplying out the first two sums gives you $mn$ terms, and therefore when you multiply by $z_1+\cdots +z_p$ you get a total of $(mn)p$.
And so on.