Calculating number of combinations of multiple sets, each containing different number of elements I'm not a math genius so please consider that when posting your explanation. 
I have the following sets, arbitrarily named:
a [a1, a2, a3]
b [b1, b2]
c [c1, c2, c3, c4]
d [d1, d2, d3, d4, d5]
e [e1, e2, e3]
f [f1, f2]
g [g1]
Question #1) What is the total number of combinations if I have to select one element from each set? I think I know this answer, but I want to confirm. I believe this is the rule of products, so the answer is the number of elements in each set multiplied together. 
In the case above, it would be:
3 * 2 * 4 * 5 * 3 * 2 * 1 = 720 combinations
Question #2) What is the total number of combinations if I have to select one element from set a, one element from set b, and one element from 4 of the 5 following sets: c, d, e, f, g?
How would this answer change if there was another set 'h' with 6 elements [h1, h2, h3, h4, h5, h6] and I had to still select one element each from set a and b, but now had to select one element each from 4 of the now 6 sets (c, d, e, f, g, h)?
Thank you for the help in advance.
 A: Your answer to the first question is correct.  
You can apply the same method to the second question.
Selections from sets a, b, c, d, e, f: $3 \cdot 2 \cdot 4 \cdot 5 \cdot 3 \cdot 2$
Selections from sets a, b, c, d, e, g: $3 \cdot 2 \cdot 4 \cdot 5 \cdot 3 \cdot 1$
Selections from sets a, b, c, d, f, g: $3 \cdot 2 \cdot 4 \cdot 5 \cdot 2 \cdot 1$
Selections from sets a, b, c, e, f, g: $3 \cdot 2 \cdot 4 \cdot 3 \cdot 2 \cdot 1$
Selections from sets a, b, d, e, f, g: $3 \cdot 2 \cdot 5 \cdot 3 \cdot 2 \cdot 1$
To find the number of selections from sets a and b and four of the five sets c, d, e, f, g, we add the above results to obtain
\begin{align*}
3 \cdot 2 \cdot & (4 \cdot 5 \cdot 3 \cdot 2 + 4 \cdot 5 \cdot 3 \cdot 1 + 4 \cdot 5 \cdot 2 \cdot 1 + 4 \cdot 3 \cdot 2 \cdot 1 + 5 \cdot3 \cdot 2 \cdot 1)\\
& = 3 \cdot 2 \cdot 4 \cdot 5 \cdot 3 \cdot 2 \cdot 1\left(\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{4}\right)
\end{align*}
The same method could be applied to your third question, but since there are 
$\binom{6}{4} = 15$ cases, calculating the answer is tedious.  The cases are selections from the sets:
a,b,c,d,e,f
a,b,c,d,e,g
a,b,c,d,e,h
a,b,c,d,f,g
a,b,c,d,f,h
a,b,c,d,g,h
a,b,c,e,f,g
a,b,c,e,f,h
a,b,c,e,g,h
a,b,c,f,g,h
a,b,d,e,f,g
a,b,d,e,f,h
a,b,d,e,g,h
a,b,d,f,g,h
a,b,e,f,g,h
