Identical vs distinct in balls to bins problems 
A committee of size $5$ is chosen from $A_1,\ldots,A_9$. Find the probability that the committee contains $A_6.$

The answer is $\frac{\binom 84}{\binom95}.$

There are $7$ churches in a town. Three visitors pick churches at random
  to attend. Find the probability that they choose $3$ different churches.

The answer is $\frac{7 \cdot 6 \cdot 5}{7^3}.$
I am interested in the numerators of both answers. In the first case we have an injective function $\{\text{identical elements}\} \to \{\text{distinct elements}\}$ and the function in the second case is $\{\text{distinct elements}\} \to \{\text{distinct elements}\}$ which is also injective.
In what sense are the elements in the domain of either function identical or distinct? What makes the three people in the second problem distinct from each other while the five people in the first problem are identical?
 A: 
A committee of size $5$ is chosen from $A_1, A_2, A_3, \ldots, A_9$.  Find the probability that the committee contains $A_6$.  

What are we doing in this problem is counting five-element subsets of a nine-element set.  
The number of ways we can select a five-element subset of a nine-element set is the number of injective functions from a set of five positions on the committee to the nine available people in which the order of selection does not matter, that is, up to a permutation of the positions on the committee.  The number of injective functions from a set of five elements to a set of nine elements is 
$$P(9, 5) = 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 = \frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4!}{4!} = \frac{9!}{4!} = \frac{9!}{(9 - 5)!}$$
Since the order of selection does not matter, we must divide by the $5!$ orders in which the same set of five people can be selected for the committee, which yields
$$\binom{9}{5} = \frac{1}{5!} \cdot \frac{9!}{(9 - 5)!} = \frac{9!}{5!(9 - 5)!}$$
For the numerator, since $A_6$ is placed on the committee, we are selecting a subset of four of the remaining eight people.  This is an injective function from the set of four remaining positions on the committee to the eight remaining people in which the order of selection does not matter.

There are $7$ churches in a town.  Three visitors pick churches at random to attend.  Find the probability they choose $3$ different churches.

Let's call the people A, B, and C.  Notice that A selecting the Anglican church, B selecting the Baptist church, and C selecting the Catholic church is different from A selecting the Catholic church, B selecting the Anglican church, and C selecting the Baptist church.  Hence, the order matters.  
The denominator $7^3$ counts the number of functions from visitors to churches, that is, the number of ways the three visitors can choose one of the seven churches to attend.
The numerator $7 \cdot 6 \cdot 5$ counts injective functions from visitors to churches, that is, the number of ways the three visitors can choose three different churches to attend.
