Combinatorics and new Bose-Einstein Statistics Problem Can you check my solution? Is this answer correct?
As there are four indistinct particles, and their sum must add 4, I am considering that exist the following partitions:
(4,0,0,0) implies C(17,1)= 17
(3,1,0,0) implies C(10,1)*C(2,1)= 20
(2,2,0,0) implies C(5,1)*C(5,1)= 25
(2,1,1,0) implies C(5,1)*C(2,1)*C(2,1)= 20
(1,1,1,1) implies C(2,1)*C(2,1)*C(2,1)*C(2,1)= 16
Finally, 17 + 20 + 25 + 20 + 16 = 98. 
Winther, who is a postdoc in theoretical astrophysics at the ICG in Portsmouth, gave me a "translation" of the problem:
Translating the problem to a more real world scenario might help: there are four wallets in a bag and each wallet contains from zero to four $1$ notes. The total money in the wallets combined is $4$. 
Your questions in terms of this analogy: "Is the energy of each particle the same?" = "Do we have the same money in all wallets?" Not necessarily. 
"some of the energy of the particles can cancel" = "some of notes cancel each other" No they can't. "Are we applying Pauli's principle?" No, bosons do not satisfy this principle.
Here is the problem:
Bose Statistics Problem

5) A physical system consists of four identical particles. The total energy of the system is $4E_0>0$. Each of the particles can have an energy level equal to $kE_0$ for $k\in\{0,1,2,3,4\}$. A particle with energy $kE_0$ can occupy any one of the $k^2+1$ distinct energy states at that energy level. How many distinct energy configurations can the system have? [In statistical mechanics the particles are said to obey Bose-Einstein statistics.]

After my "research" in many physics websites, I found this:

Consider an energy level $\varepsilon_i$ with degeneracy $g_i$, containing $n_i$ bosons. The states may be represented by $g_i-1$ lines, and the bosons by $n_i$ circles; distinguishable microstates correspond to different orderings of the lines and circles. For example, with $9$ particles in the $8$ states corresponding to a particular energy, a particular microstate might be:
  
The number of distinct orderings of lines and circles is:
  $$t_i = \frac{(n_i+g_i-1)!}{n_i!(g_i-1)!}.$$ 
A particular distribution has a specified number of particle $n_i$ within each of the possible energy levels $\varepsilon_i$. The total number of microstates for a given distribution is therefore:
  $$t(\{n_i\}) = \prod_i \frac{(n_i+g_i-1)!}{n_i!(g_i-1)!}.\tag 8$$

I also found a couple of Bose-Einstein Statistics problems:
A combinatorics problem related to Bose-Einstein statistics
Indistinguishable particles: obeing "Bose-Einstein Statistics"
 A: This is a guide of what you have regarding the binomial coefficient $t_i$.
The binomial in $t_i$ is basically the problem of arranging a word of $n_i+g_i-1$ letters, from which $n_i$ letters are, say A (or bosons $b$), and $g_i-1$ letters are, say B (or boxes $b$). What you have then is a Mississippi problem (very well worked out in the early part of combinatorics courses and books). Martin's book Counting: the art of enumerative combinatorics can help you to understand these concepts.
Now to the problem. You can form $N!$ words from $N$ distinct letters. Consider now that the letters are not all different, and in fact that you have only two types of letters, say $n_A$ $A$'s and $n_b$ $B$'s such that $n_A+n_B=N$. Then $N!$ is an overcounting of all the possible distinct words that you can form (try this with, say $3$ $A$'s and $2$ $B$'s so that you can convince yourself of this fact). By the division principle, the total number of words of length $N$ that you can form with  $n_A$ $A$'s and $n_b$ $B$'s is
$$
\frac{(n_A+n_B)!}{n_A!n_B!}.
$$
Now it is easy to say that there are $n_A$ bosons and $n_B+1$ boxes (you need essentially $n_B$ "bars" to form $n_b+1$ boxes).
