Counting positive integral solutions to an equation (3) This site gives me lot of opportunity to learn mathematics. With my previous question, I am clear in handling questions like finding number of integral solutiuons to

$y_1+y_2+y_3+y_4+y_5+y_6=33$
where $1≤y_i≤8$ for each $i=1,2,3,4,5,6$

Thanks to @N. F. Taussig for the clear answer provided. This is all I have in syllabus. But, while practicing similar question, I got this doubt.

Suppose sum like

$1+1+1+1+1+28=33$ 
$28+1+1+1+1+1=33$

are same.

(1)What is the best combinatorial methods for such problems?
(2) My previous question referred is often treated as distribution of identical objects into distinct bins. If so, are these new problems same as distribution of identical objects into identical bins?
Note: I spend a lot of time now for studying the approach. I initially thought there will be some easier methods like starts and bars. But could not find any or methods explained are going over my head. Initially thought this is more simple.
Please guide me. Thanks
 A: Your comment about identical bins is exactly correct; you no longer wish to distinguish the summands (the $y_i$) from each other.
The number of ways to write
$$n = \sum k_i$$
with all $k_i > 0$, and all summands indistinguishable, is known as the number of partitions of $n$.
This is a huge topic by itself, and I have to point you to the Wikipedia page on partitions. 
You are specifically asking for the number of partitions of 33 with exactly 6 parts. The number of partitions of $n$ with $k$ parts ($p_k(n)$) satisfies a recurrence:
$$p_k(n) = p_k(n − k) + p_{k−1}(n − 1).$$
We have some easy base cases: $p_k(k) = 1$ (there is 1 way to write $k$ as a sum of $k$ parts, namely $1 + 1 + \cdots + 1$), and $p_1(n) = 1$. It is also not too hard to see that $p_2(n) = \lfloor n/2 \rfloor$; beyond that, I refer you again to the Wikipedia page and its references.
Rather than working through this to find $p_6(33)$ for you, which is tedious, I will prove that the recurrence holds.

Lemma: 
$$p_k(n) = p_k(n − k) + p_{k−1}(n − 1).$$
Proof:
Take an arbitrary partition of $n$ into $k$ parts. Either (a) all parts are at least $2$, or else (b) they are not. 
(a): If all $k$ parts are at least $2$, then we may subtract $1$ from each part and obtain a partition of $n-k$ with $k$ parts; similarly, if we have a partition of $n-k$ into $k$ parts, we can add $1$ to each part and obtain a partition of $n$ into $k$ parts of size at least $2$, so the partitions of $n$ into $k$ parts of size at least $2$ are in bijection with the partitions of $n-k$ into $k$ parts.
(b): If there is a part with size 1, throw it away to obtain a partition of $n-1$ into $k-1$ parts. Similarly, to any partition of $n-1$ into $k-1$ parts, we can add a part of size $1$ to obtain a partition of $n$ into $k$ parts, so these are also in bijection.
Since these two cases are mutually exclusive and cover all possibilities, this proves the recurrence, which you can now apply to your problem.

If you are interested in a more systematic (but lengthy) approach to this general topic, I refer you further to the twelvefold way. This includes some very challenging material, but it appears you're headed toward it on your own, and it might be a useful resource for you later.
