Advanced Math Help with Any Problems!

Simplification of expression (related to multinomial theorem)
Is there a simplified way to write the following:
$$\underset{r_1+r_2+\cdots <!--\; \cdots \; -->+r_n=t,r_k\in <!--\; \in \; -->\mathbb{N}}{\sum <!--\; \sum \; -->}\frac{t!}{r_1!r_2!\cdots <!--\; \cdots \; -->r_n!}\underset{k=1}{\overset{n}{\prod <!--\; \prod \; -->}}f(k{)}^{r_k}$$
This is very similar to the multinomial theorem formula, however instead of $r_k$ being non-negative, $r_k$ is a natural number. In an expansion this is equivalent to just taking the values where every term is exponentiated to a power of one or greater, for example in the expansion of $(a+b+c{)}^4$ just taking $3(2a^{2}bc+2a{b}^{2}c+2ab{c}^2)$.

Set Theory - I'm new at set theory and this problem I don't know how to start it please help
let's consider this two set A and B as:
$$A=\{\frac{(\mathrm{\Pi <!--\; \Pi \; -->})}{6}+\frac{k\mathrm{\Pi <!--\; \Pi \; -->}}{3}|k\in <!--\; \in \; -->Z\}$$
$$B=\{-<!--\; -\; -->\frac{(\mathrm{\Pi <!--\; \Pi \; -->})}{4}+\frac{k\mathrm{\Pi <!--\; \Pi \; -->}}{3}|k\in <!--\; \in \; -->Z\}$$
I started learning mathematics on my own, in my house, through the internet, I stadied logic, and now Set Theory, and I encountered some problems like , this set builder have this numbers, and this other set have this numbers, what is the result if you do intersection or union between them or something like this.
and this question I encountered it I answered the first one about how do you answer if I asked you is this two sets have in their intersection an emptyset like: $A\cap <!--\; \cap \; -->B=\mathrm{\varnothing <!--\; \varnothing \; -->}$ and after that I had this question below Question is:
How to determine the set:
$$(A\cup <!--\; \cup \; -->B)\cap <!--\; \cap \; -->[-<!--\; -\; -->\frac{\mathrm{\Pi <!--\; \Pi \; -->}}{2};\frac{\mathrm{\Pi <!--\; \Pi \; -->}}{2}]$$
in my first question I answered it using (A $\cap <!--\; \cap \; -->$ B is x in A and x in B ... and I said that means the first equation $\mathrm{\pi <!--\; \pi \; -->}/6+k\mathrm{\pi <!--\; \pi \; -->}/3=(-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}/4+k\mathrm{\pi <!--\; \pi \; -->}/3))$ and I solved that equation and I found that $k-<!--\; -\; -->k^{\prime }=a/b$ which is contradiction so the intersection is an empty set, now the second question is what I wrote above I still can't solve it.
here the only question that I need answer for is the one about this :
(
How to determine the set:
$$(A\cup <!--\; \cup \; -->B)\cap <!--\; \cap \; -->[-<!--\; -\; -->\frac{\mathrm{\Pi <!--\; \Pi \; -->}}{2};\frac{\mathrm{\Pi <!--\; \Pi \; -->}}{2}]$$
)<<< this is the one

Information about this type of recurrence relation
I've been noodling with a problem that requires me to find a set of recurrence relations and their solutions. In particular, all the recurrence relations are of the form:
$$u_{n+1}=u_n+p(n)$$
where p(x) is a polynomial.

Find a closed form for $\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}{k}^5$ using generating functions.

Equivalence relation problem - discrete math
I'm having issues trying to start this question, what does defining a new relation mean?
The question:
Suppose $\mathrm{\rho <!--\; \rho \; -->}$ is reflexive and transitive relation on a set S. Define a new relation $\mathrm{\sigma <!--\; \sigma \; -->}$ as follows: $a\mathrm{\sigma <!--\; \sigma \; -->}b$ is true if and only if both $a\mathrm{\rho <!--\; \rho \; -->}b$ and $b\mathrm{\rho <!--\; \rho \; -->}a$ are true. Show that $\mathrm{\sigma <!--\; \sigma \; -->}$ is an equivalence relation.

Discrete math surjective function proof
Let E,F and G be given sets and let the functions $f:E\to <!--\; \to \; -->F$ and $g:F\to <!--\; \to \; -->G$ be given. Consider the claim:
if f is surjective and g is surjective $⟹<!--\; ⟹\; -->g\circ <!--\; \circ \; -->f$ is surjective.
The claim is either true or false. If it is true, prove it. If it is false, give example of 3 sets of E,F,G and two functions f,g where the fuctions $f:E\to <!--\; \to \; -->F$ and $g:F\to <!--\; \to \; -->G$ are surjective but $g\circ <!--\; \circ \; -->f$ isn't surjective.
How should I think about all of this? We've had so few tasks in our course on this subject

A 10-foot-long chain weighs 25 lbs. And hangs from a ceiling. Calculate the work done in raising the lower end of the chain to the ceiling so that it is at the same level as the upper end.

Calculate cardinality of a subset of all the relations above N
Define X to be set of all the relations R over N such that $R^{\ast <!--\; \ast \; -->}=\mathbb{N}\times <!--\; \times \; -->\mathbb{N}$, where R∗ is the transitive closure of R.
What is the cardinality of the set X?
What I've tried so far:
I notice that all the transitive relations besides $\mathbb{N}\times <!--\; \times \; -->\mathbb{N}$ are not in X. And I know that the cardinality of all the transitive relations over N is c.

Question: $(x_n\to <!--\; \to \; -->x⟹<!--\; ⟹\; -->\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim sup}f(x_n)\le <!--\; \le \; -->f(x))⟹<!--\; ⟹\; -->\{x\in <!--\; \in \; -->X|f(x)<c\}\text{is open for any}c\in <!--\; \in \; -->R?$

Discrete logarithm problem: does the base have to be a generator of Z?
${\mathbb{Z}}_p$ under multiplication, $p=13,Z=\{1,\dots 12\}$ I need to find the discrete logarithm of base 5 to 8:
So essentially $5^x\equiv <!--\; \equiv \; -->8(\mathrm{mod}13)$
However, In this problem 5 is not a generator of ${\mathbb{Z}}_{13\cdot <!--\; \cdot \; -->}$ 5 produces a cyclical group {5,12,8,1} which does not contain all the elements of ${\mathbb{Z}}_{13\cdot <!--\; \cdot \; -->}$
I understand there is a solution to this, but can someone clarify this rule for me?

Combinatorics problem involving teams (Discrete math)
A company has 10 men and 18 women. A work team consists of two workers. What is the maximum number of work teams (man-man,woman-woman,man-woman) that can be formed from this group? How many different ways can this maximum number of work teams be formed? I know the max amount of teams is 14. I tried 28 choose 2 for the maximum number of team combinations but it's wrong.

The relation R is an equivalence relation on the set A. Find the distinct equivalence classes of R.
$A=\{0,1,2,3,4\}$
$R=\{(0,0),(0,4),(1,1),(1,3),(2,2),(3,1),(3,3),(4,0),(4,4)\}$
Here is the solution:
$[0]=\{x\in <!--\; \in \; -->A|xR0\}=\{0,4\}$
$[1]=\{x\in <!--\; \in \; -->A|xR1\}=\{1,3\}$
$[2]=\{x\in <!--\; \in \; -->A|xR2\}=\{2\}$
$[3]=\{x\in <!--\; \in \; -->A|xR3\}=\{1,3\}$
$[4]=\{x\in <!--\; \in \; -->A|xR4\}=\{0,4\}$
Note that $[0]=[4]$ and $[1]=[3]$. Thus the distinct equivalence classes of the relation are {0,4}, {1,3}, and {2}.
My problem here is that I am not understanding the solution. I do not understand how it came up with an answer for each equivalence class of every element A. As in, how is {0,4} equal to $\{x\in <!--\; \in \; -->A|xR0\}$, and how is that equal to [0]? I can understand that $[0]=[4]$ since they both equal {0,4} but I'm not sure how to arrive at that answer.
I'm trying this problem:
$A=\{a,b,c,d\}$
$$R=\{(a,a),(b,b),(b,d),(c,c),(d,b),(d,d)\}$$
However, I am lost because I do not understand how to arrive at answers for every element in A.

Discrete maths
$$\begin{gathered} -<!--\; -\; -->3a_n+a_{n+1}=2^n \\ {a}_0=0 \end{gathered}$$

Discrete Math Proof: Divisibility equivalence
For all integers a, b, d, if d divides a, and d divides b, then d divides $(3a+2b)$ and d divides $(2a+b)$. Prove the statement.
What Assumptions do I need to make at the beginning of this proof that include $(3a+2b)$ and $(2a+b)$. I can start off the proof with:
Suppose a, b, d are integers and that d divides a, and d divides b. Then by definition of divisibility, there exist integers c, k, such that $a=dc$ and $b=dk$.

Derive a homogenous linear recurrence from $x_n=2^n+F_n$
Suppose that the sequence $x_n$ satisfies that
$$x_n=2^n+F_n$$
where $\{F_n\}$ denotes the Fibonacci sequence, I want to derive a homogenous linear recurrence for $x_n$
I can show that
$$x_n-<!--\; -\; -->x_{n-<!--\; -\; -->1}-<!--\; -\; -->x_{n-<!--\; -\; -->2}=\frac{1}{4}{2}^n$$
By plug in the relationship of Fib Sequence, however the above is non-homogenous, can anyone help?

Discrete Math on Induction proofs.
Show by induction that the following formulas hold for
$$\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{i}^3=\frac{n^2(n+1{)}^2}{4}$$
Not sure how to go about this problem. Can someone help please?

Discrete Math - Combinatorics - Trinomial Coefficients question
Let $k,l,m,n\in <!--\; \in \; -->Z\ge <!--\; \ge \; -->0$ be such that $n=k+l+m$. The trinomial coefficient $(\genfrac{}{}{0ex}{}{n}{k,l,m})$ is given by the rules:
1. for $k+l=n$, $(\genfrac{}{}{0ex}{}{n}{k,l,0})=(\genfrac{}{}{0ex}{}{n}{k,0,l})=(\genfrac{}{}{0ex}{}{n}{0,k,l})=(\genfrac{}{}{0ex}{}{n}{k})$
2. $(\genfrac{}{}{0ex}{}{n}{k,l,m})=(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->1}{k-<!--\; -\; -->1,l,m})+(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->1}{k,l-<!--\; -\; -->1,m})+(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->1}{k,l,m-<!--\; -\; -->1})$
The following questions use this definition.
(a) What are all the trinomial coefficients for $n=1,2,3$?
(b) Describe the "triangle" of trinomial coefficients (Hint: Think three dimensional Pascal's triangle).
I don't understand the notation at all: n choose k,l,m. What does that even mean? And I don't get how to set up part (a) at all either. It's just very confusing. For (b) I can visualize a Pascal's "pyramid." One where the number below is a sum of the 3 above it. Something like that. But other than that, I'm not really sure what's going on.

Discrete Math: Multiplying a set by $\mathrm{\varnothing <!--\; \varnothing \; -->}$. How would you multiply any set by $\mathrm{\varnothing <!--\; \varnothing \; -->}$? Lets say $A\times <!--\; \times \; -->\mathrm{\varnothing <!--\; \varnothing \; -->}$. Would that simply be equal to $\mathrm{\varnothing <!--\; \varnothing \; -->}$? or Would I write out $(a,\mathrm{\varnothing <!--\; \varnothing \; -->}),(a_1,\mathrm{\varnothing <!--\; \varnothing \; -->}),(a_2,\mathrm{\varnothing <!--\; \varnothing \; -->})$ etc. If possible could you explain your answer in words or short proof?

I have a function $f(x)=g(x)h(x)$ . I have either of the following two conditions to hold
g(x) has positive value and it is increasing and h(x) has negative values but is decreasing.
g(x) has positive value and it is increasing and h(x) has negative values but is increasing.
Can I say that f(x) is a quasiconcave function?

Discrete Math Clause Count Question
In a CNF formula, a clause contains one or more terms. Each term is either a variable, or the negation of a variable. The only limitations we will place on the clauses (for this problem) is that if a clause mentions a variable (or its negation) once, then it must not mention the same variable (or its negation) again. Therefore, both of the following clauses are disallowed: $(a\vee b\vee a)(a\vee b\vee \neg a)$ However, your program might well generate similar clauses with different negation patterns. The following are four different clauses: $(a\vee b),(a\vee \neg b),(\neg a\vee b),(\neg a\vee \neg b)$
Part (a) - In general, as your program runs, it will generate larger and larger clauses, with a maximum of v terms per clause, if there were v variables mentioned in the original formula. Calculate the number of possible clauses which have exactly n terms, if the original formula mentioned v variables
Part (b) - Calculate the sum total number of clauses (of various sizes) that your program might generate, if the original formula mentioned v variables.
Part (c) - Suppose the original formula mentioned v variables, but every single clause had two or fewer terms. Calculate the total number of clauses which your program might generate.
Part (d) Based on your result from Part (b), give a very rough (thousands, millions, etc) estimate of how many clauses your program might generate if $v=10$, and the clauses in the original formula can be of any size. What would happen if v were to double? How does the count change?