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
My common use case is large sets (thousands or millions of elements) but only a handful of links.
For me a clustered recursion seems to be a good algorithm. The idea is simple but it is a bit hard for me to explain it clearly. I’ll hope you’ll understand it from the below. An input for a clustered recursion is the following. Assume that we are given sets $S_1,\dots, S_n$. Then for each $i$ and each subset $I$ of $\{i+1,\dots, n\}$ we count a number $(i,I)$ of elements of $S_i$ which linked to exactly members of $I$. Also assume that we are given a subset $Z$ of $\{1,\dots, n\}$ of indices $i$ of sets $S_i$ which do not need links to be listed. Clearly, suffices to consider reduced inputs $(i,I)$ with $I\cap Z\varnothing$, with $(i,I):=\sum \{(i,I’):I\subset I’\subset I\cup Z\}$. For instance, in your example we have $n=3$, $Z=\{1\}$, $(1,\{2\})=1$, $(1,\{3\})=2$, $(1,\varnothing)=1$, $(2,\varnothing)=3$, $(3,\varnothing)=2$, and all other $(i,I)$ are zero. Then using the same arguments as in the question, we can count the total number $N$ of possibilities as follows. For each subset $I$ of $\{2,\dots, n\}$ we recursively calculate the total number $N_I$ of possibilities in which the first element is linked with exactly members of $I$ and then count $N=\sum \{N_I: I\subset \{2,\dots, n\}\}$. For instance, in your example there are
$(1,\{2\})=1$ choice to choose an element linked exactly with the set $S_2$. In this case we add $2$ to and remove $1$ from the set $Z$ of indices of sets which do not need links to be listed. This gives us the same problem, but with the smaller number of sets. Since the elements of the set $S_2$ are not linked with other sets, in this case we have only $(2,\varnothing)=3$ choices. So in total we have $N_{\{2\}}=(1,\{2\})(2,\varnothing)=3$.
$(1,\{3\})=2$ choices to choose an element linked exactly with the set $S_3$. In each of these cases we add $3$ to and remove $1$ from the set $Z$. This gives us the same problem, but with the smaller number of sets. Since the elements of the set $S_3$ are not linked with other sets, in this case we have only $(3,\varnothing)=2$ choices. So in total we have $N_{\{3\}}=(1,\{3\})(3,\varnothing)=4$.
$(1, \varnothing)=1$ choice to choose an element not linked with any set $S_i$. In this case we remove $1$ from the set $Z$, but add to it nothing. So in total we have $N_\varnothing =(1,\varnothing)=1$.
Thus we have $N= N_\varnothing + N_{\{2\}}+ N_{\{3\}}=1+3+4=8$.
Best Answer
A "pairing" is an injective function from $\{1,2,3\}$ to $\{4,5,6\}$. For the image of $1$ you have $3$ choices. There are two choices left for the image of $2$, and one choice for the image of $3$. The conclusion comes from the product rule: there are $1\times 2\times 3 = 6$ injective functions.