How does stars and bars apply in this problem? The problem is from a mock contest written by AoPS user Stormerstyle.

$n$ dice, labeled $d_1$, $d_2$...$d_n$, are simultaneously rolled. For a given positive integer $n$, let $P_n$ be the number of ways the $n$ dice can be rolled such that for all integers $i$ where $1\le i\le n-1$, the number rolled on $d_i$ is greater than or equal to the number rolled on $d_{i+1}$. Let $k$ be $P_3+P_4+...+P_{70}$. Compute the sum of the three largest prime factors of $k+28$.

It is not explicitly given, but from working on a solution that fits well with the actual solution, it is clear that the dice have 6 sides. Essentially, there are $n$ dice that are simultaneously, and $P_n$ denotes the number of ways that the dice (and thus the integers from 1 to 6) can be set, such that $d_n \ge d_{n-1} \ge \dots d_1$, if we let $d_i$ denote the number shown on the die instead of the die object itself. 
My attempt consisted of finding $P_3 = 56$ and $P_4 = 126$, where I then found that $P_k$ and $P_{k+1}$ are closely related, but I couldn't generalize to more than two $P_i$, because nothing canceled out well and I wasn't able to formalize it. Basically, it was a dead end.
One solution given was using stars and bars and the hockey stick identity by noticing that $P_i = \dbinom{i+5}{i}$, then simply adding from $P_3$ to $P_{70}$ and then using the hockey stick identity at the end. I am unsure of how to show the identity $P_i = \dbinom{i+5}{i}$. I do not understand why we are using $i+5$ for the top of the binomial coefficient. I noticed that my $P_3$ and $P_4$ values fit the stars and bars solution, but I still do not understand why. My question is, why is this true for $P_i$?
 A: We place $i + 5$ empty spots next to each other. Now pick a subset of five bars from these spots. These partition our $i$ remaining empty spots into six parts (some of them might be empty). Each way to pick these five bars, exactly prescribes a solution: the first part of remaining empty spots has all dices that are 1, the second has dices that are 2, et cetera. Also any solution for the dices can be uniquely determined by 5 bars in the other setting.
Hence, the number of ways these dices can be rolled in increasing fashion, is equal to the amount of ways you can pick 5 bars from a set of $i+5$ items. Which is ${i+5 \choose 5} = {i+5 \choose i}$.
A: Put $P(n,m)$ be the number of ways to throw the $n$ dice, in a non decreasing  sequence ending with $m$.
Then clearly $P(n+1,m)=P(n,1)+P(n,2) + \cdots + P(n,m)$.
Alternatively consider that
$$
\eqalign{
  & {\rm N}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ {1\, \le \,d_{\,1}  \le d_{\,2}  \le  \cdots  \le d_{\,n} \, \le \,6} \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad {\rm N}{\rm .}\,\,{\rm sol}{\rm .}\;\left\{ {0\, \le \,x_{\,1}  \le x_{\,2}  \le  \cdots  \le x_{\,n} \, \le \,5} \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad {\rm N}{\rm .}\,\,{\rm sol}{\rm .}\;\left\{ \matrix{
  0 \le y_{\,k}  \le 5 \hfill \cr 
  \,y_{\,1}  + y_{\,2}  +  \cdots  + y_{\,n} \, \le \,5 \hfill \cr}  \right. \cr} 
$$
The latter has to do with "stars and bars", or better with  the compositions
of the integers from $0$ to $5$, into $n$ parts.
So
$$
N = \sum\limits_{0 \le k \le 5} {\left( \matrix{
  k + n - 1 \cr 
  k \cr}  \right)}  = \sum\limits_{0 \le k \le 5} {\left( \matrix{
  5 - k \cr 
  5 - k \cr}  \right)\left( \matrix{
  k + n - 1 \cr 
  k \cr}  \right)}  = \left( \matrix{
  5 + n \cr 
  5 \cr}  \right)
$$
