I need to find the combinatory formula for this set. 
Problem 1. Fix a positive integer $n$. For every integer $S \geq n$, let $N_{n,S}$ denote the number of possible ways in which a sum of $S$ can be obtained when $n$ dice are rolled. For example, for $n = 3$ dice and a sum $S = 5$, we have $N_{3,5} = 6$, counting the following possible triples:
  \begin{align}
\left(3,1,1\right), \quad \left(1,3,1\right), \quad \left(1,1,3\right), \quad
\left(2,2,1\right), \quad \left(1,2,2\right), \quad \left(2,1,2\right) .
\end{align}
(a) Consider the sets
  \begin{align}
A = \left\{ \left(a_1, a_2, \ldots, a_n\right) \in \mathbb{Z}^n ; \  a_i \geq 1 \text{ for all } i, \text{ and } \sum_{i=1}^{n} a_i = S \right\}
\end{align}
  and
  \begin{align}
A_j = \left\{ \left(a_1, a_2, \ldots, a_n\right) \in \mathbb{Z}^n ; \  a_i \geq 1 \text{ for all } i, \ a_j \geq 7 \text{ and } \sum_{i=1}^{n} a_i = S \right\}
\end{align}
  for a fixed $j \in \left\{1, 2, \ldots, n\right\}$.
(i) Write formulas for the numbers of elements in $A$ and $A_j$, respectively. Justify your answers.
(ii) State the Inclusion-Exclusion Formula and use it to prove:
  \begin{align}
N_{n,s} = \sum_{k=0}^n \left(-1\right)^k C^n_k C^{S-1-6k}_{n-1}
\end{align}
  (where $C^a_b$ stands for the binomial coefficient $\dbinom{a}{b}$).

This is the problem I am having difficulty with. For $A$, I think the formula is $$\left|A\right| = {S-1 \choose S-n}.$$
I can't seem to figure a formula for $A_j$, and I don't understand where the variable "x" came from. I have a good understanding of the Inclusion Exclusion formula and think I could complete the proof if I had $A_j$.
 A: The  nodes of  the poset  for  use with  PIE consists  of the  subsets
$Q\subseteq [n]$ representing $n$-tuples of positive integers that sum
to $S$  where the elements at  positions $q\in Q$ are  at least seven,
plus possibly some  others. The weight of each  tuple is $(-1)^{|Q|}.$
Note that  tuples where all  elements are  less than seven  only occur
when $Q =  \emptyset$ and hence have weight  $(-1)^{|\emptyset|} = 1.$
On the  other hand tuples  with values at  least seven exactly  at the
positions  of  some  $P\subseteq  [n]$   are  included  at  all  nodes
$Q\subseteq P$ for a total weight of 
$$\sum_{Q\subseteq P} (-1)^{|Q|} =
\sum_{p=0}^{|P|} {|P|\choose p} (-1)^p
= 0$$
i.e.   zero. So  the only  contribution  comes from  the tuples  being
counted            by           $N_{n,S}.$            We           use
stars-and-bars
to compute the  cardinality of the set of tuples  being represented by
$Q.$ For stars-and-bars  the summands to the sum  are non-negative and 
hence to reduce $Q$ to   stars-and-bars we have to subtract seven from
$S$ at  each element  of $Q$  and one  at those  not in  $Q$ to  get a
standard stars-and-bars  scenario. (This is  the data that  is present
between bars no matter what.) This yields with $|Q|=k$
$${S - 7\times k - 1\times (n-k) + n-1\choose n-1}
= {S - 1 - 6k\choose n-1}.$$
We then have by PIE
$$N_{n,S} =
\sum_{k=0}^n {n\choose k} (-1)^k {S - 1 - 6k\choose n-1}.$$
To verify this we may use the closed form
$$[z^S] (z+\cdots+z^6)^n
= [z^S] z^n (1+\cdots+z^5)^n
= [z^{S-n}] \frac{(1-z^6)^n}{(1-z)^n}
\\ = [z^{S-n}] \sum_{k=0}^n {n\choose k} (-1)^k z^{6k}
\frac{1}{(1-z)^n}
\\ = \sum_{k=0}^n {n\choose k} (-1)^k [z^{S-n-6k}]
\frac{1}{(1-z)^n}
\\ = \sum_{k=0}^n {n\choose k} (-1)^k {S-n-6k+n-1\choose n-1}
= \sum_{k=0}^n {n\choose k} (-1)^k {S-1-6k\choose n-1}.$$
A: You have correctly calculated $|A|$.  However, in order to obtain the desired formula for $N_{n,S}$, it would be more useful to write $|A|$ in the form
$$|A| = \binom{S - 1}{n - 1}$$
which can be obtained from your formula by observing that 
$$|A| = \binom{S - 1}{S - n} = \binom{S - 1}{S - 1 - (S - n)} = \binom{S - 1}{n - 1}$$
Alternatively, since 
$$A = \left\{(a_1, a_2, \ldots, a_n) \in \mathbb{Z}^n \mid a_i \geq 1~\text{for all $i$ and}~\sum_{i = 1}^{n} a_i = S\right\}$$
we need to count the number of solutions of the equation
$$\sum_{i = 1}^{n} a_i = a_1 + a_2 + \cdots + a_n = S$$
in the positive integers.  A particular solution corresponds to the placement of $n - 1$ addition signs in the $n - 1$ spaces between successive ones in a row of $n$ ones.  For instance, in the case $S = 5$ and $n = 3$, 
$$1 \square 1 \square 1 \square 1 \square 1$$
choosing to fill the third, fourth, and fifth spaces with addition signs yields
$$1 1 1 + 1 + 1$$
which corresponds to the solution $(a_1, a_2, a_3) = (3, 1, 1)$.  The number of such solutions is the number of ways we can select $n - 1$ of the $S - 1$ spaces between successive ones to fill with addition signs, which is 
$$\binom{S - 1}{n - 1}$$
Since we are interested in the number of ways the sum $S$ can be obtained when $n$ six-sided dice are rolled, we must exclude those solutions in which $a_j \geq 7$ for some $j \in \{1, 2, \ldots, n\}$.  If we let 
$$A_j = \left\{(a_1, a_2, \ldots, a_n) \in \mathbb{Z}^n \mid a_i \geq 1~\text{for all}~i, a_j \geq 7,~\text{and}~\sum_{i = 1}^{n} a_i = S\right\}$$
for some fixed $j \in \{1, 2, \ldots, n\}$, then $|A_j|$ is the number of positive integer solutions of the equation 
$$\sum_{i = 1}^{n} a_i = a_1 + a_2 + \cdots + a_n = S$$
in which $a_j \geq 7$. Since $a_j \geq 7$, $a_j' = a_j - 6$ is a positive integer.  Substituting $a_j' + 6$ for $a_j$ in the equation
$$a_1 + a_2 + \cdots + a_j + \cdots + a_n = S$$
yields
\begin{align*}
a_1 + a_2 + \cdots + a_j' + 6 + \cdots + a_n & = S\\
a_1 + a_2 + \cdots + a_j' + \cdots + a_n & = S - 6
\end{align*}
which is an equation in the positive integers with 
$$\binom{S - 6 - 1}{n - 1} = \binom{S - 1 - 6}{n - 1}$$
solutions.  Hence, 
$$|A_j| = \binom{S - 1 - 6}{n - 1}$$
From there, you can use the Inclusion-Exclusion Principle to obtain the formula for $N_{n, S}$.
