Probability of certain permutations in a set 
1. Consider a uniformly random permutation of the set $\{1, 2, . . . , 77\}$.  Define the event $A =$ “in the permutation, both $8$ and $4$ are to the left of 
  $3$”.
What is Pr(A)?  
Answer: $1/3$ 

I am trying to understand why this the answer and the question somewhat. I can understand that the possibilities for the permutation would be
$8-4-3$ and,
$4-8-3$.
Since there are $2$ such permutations, we have $Pr(A) = \frac{2}{3!} = 1/3$.  
Why are we dividing by $3!$ in $\frac{2}{3!}$ ?
The set has $77$ elements, so shouldn't the denominator be $77!$? When I read this question, there would be $1$ spot in the random $77$-element permutation where there is a $3$, then everything to the left of that would contain an $8$ and $3$, but they don't necessairly have to be next to eachother as long as they are to the left of $3$. Since it's a random permutation of $77$ elements, what if $3$ lands as the first element? $8$ and $4$ would have no where to go.

2. Let $n \ge 2$ be an integer. Consider permutations $a_i,a_2,...,a_n$ of the set $\{1,2,..., n\}$ for which $a_1 < a_2$.  
How many such permutations are there? 
Answer: $\frac{n!}{2}$

Here is my intuition. I would like to know how to arrive to this answer.
The first two elements can be  
$(1,k), 2 \le k \le 6 \implies$ 5 ways, or
$(2,k), 3 \le k \le 6 \implies$ 4 ways, or
$(3,k), 4 \le k \le 6 \implies$ 3 ways, or
$(4,k), 5 \le k \le 6 \implies$ 2 ways, or
$(5,6) \implies$ 1 way.  
This comes out to $5!$, but $5!$ also counts the pairs the other way around, so we divide by $2$ to get   
$\frac{5!}{2} = \frac{(n-1)}{2}$
Why is it $\frac{n!}{2}$?
 A: Yes, there are 77 numbers in the set so 77! permutations- but that is not at all relevant!  Instead look at "3", "4", and "8".  There are 3!= 6 permutations of those 3 numbers only.  We can list them: 348, 384, 438, 483, 834, and 843.  In two of them, 483 and 843, a proportion of 1/3, have "4 and 8 before 3".  In all 77! of the permutations, all possible permutations of "348" also appear with "3 and 4 appear before 3" in 1/3 of them.
A: For question 1, you uniformly draw a permutation of the numbers $\{1,2,\ldots,77\}$. As you write, the order of the numbers $3$, $4$, and $8$ in this permutation will be one of the following $3!$ options.
$$3, 4, 8$$
$$3, 8, 4$$
$$4, 3, 8$$
$$4, 8, 3$$
$$8, 3, 4$$
$$8, 4, 3$$
As our choice is uniform, each possibility is equally likely, and since exactly two of them has both $8$ and $4$ to the left of $3$, you get $P(A) = \frac{2}{3!} = \frac{1}{3}$.
For question 2, it looks like you are not counting the permutations correctly. For instance, if $n=6$, and if $a_1$ occupies the $5$th position, then $a_2$ has to occupy the $6$th position - but there is not just one permutation that fulfills this, there are $4!$, since the first four positions can be permuted at will.
The answer is $\frac{n!}{2}$, because in any of the $n!$ permutations of $\{1,2,\ldots,n\}$, we have either $a_1<a_2$ or $a_2<a_1$, and because of uniformity, exactly half of them will have $a_1<a_2$.
