Advanced Math Help with Any Problems!

Question: Provide further motivation for defining $p\to <!--\; \to \; -->q$ to be true when p is false For the first change, we call the resulting operator imp1.
Show that p imp1 q logically equivalent q imp1 p.
$$\begin{array}{|lll|}\hline p& q& pimp1\\ T& T& T\\ T& F& F\\ F& T& F\\ F& F& T\\ \hline\end{array}$$

Expressing statements in Discrete math
Given that
A is the set of all Alpha's
M is the set of all Men
how do I express this statement: Not all Alpha's are Men
.............
My attempt:
$A\subset <!--\; \subset \; -->S=0$
in other words saying that A is not a subset of S, but I can't use the not subset symbol on this problem.

Question about getting a formula for a recurrence relations
So basically, I was watching video above which is on recurrence relations and I had a question about this statement:
$$a_n=a_{n-<!--\; -\; -->1}+6a_{n-<!--\; -\; -->2}$$
I understand how he got the $(-<!--\; -\; -->2{)}^n$ and $(3{)}^n$, but not about how he is adding them and then multiplying them by the variables $\mathrm{\alpha <!--\; \alpha \; -->}$ and $\mathrm{\beta <!--\; \beta \; -->}$. He said that there is a proof online about why there is always going to be an $\mathrm{\alpha <!--\; \alpha \; -->}$ and a $\mathrm{\beta <!--\; \beta \; -->}$ that will always make this statement true, but I wasn't able to find it and I was hoping that somebody could give me a step by step explanation about how this is. I was also wondering about the general case for getting a formula for recurrence relations in this form.
Note: I read somewhere that you can derive this from generating functions, but I don't have a strong background in them, so I was wondering if there is another way to derive the relation.

Need help understanding $\mathrm{\exists <!--\; \exists \; -->}x\mathrm{\forall <!--\; \forall \; -->}yvs\mathrm{\forall <!--\; \forall \; -->}x\mathrm{\exists <!--\; \exists \; -->}y$
My understanding is that for $\mathrm{\exists <!--\; \exists \; -->}x\mathrm{\forall <!--\; \forall \; -->}y$, there can only be one x value that is true for every single y value. Meaning theres only one x value (which cannot be changed) for every single different y value. The statement $\mathrm{\exists <!--\; \exists \; -->}x\mathrm{\forall <!--\; \forall \; -->}y(p(x,y))$ is true when there is one x value (lets say $x=0$) that is true for $y=-<!--\; -\; -->2,-<!--\; -\; -->1,0,1,2$,... (for every single y). Correct me if I am wrong but this is my understanding of this notation.
And now my understanding for the second notation $\mathrm{\forall <!--\; \forall \; -->}x\mathrm{\exists <!--\; \exists \; -->}y(p(x,y))$ is that for every x value, there exists a y such that p(x,y). Meaning for every x value ($x=-<!--\; -\; -->2,-<!--\; -\; -->1,0,1,2,...$) there can be a different y value for each x value so that the statement is true.
I dont really know how to explain this well but I'll try to summarize my understanding. If the notation is $\mathrm{\exists <!--\; \exists \; -->}x\mathrm{\forall <!--\; \forall \; -->}y$ then theres only one x that cannot be changed that is true for every y. If the notation is $\mathrm{\forall <!--\; \forall \; -->}x\mathrm{\exists <!--\; \exists \; -->}y$ then the y value doesnt have to be the same y value for every x value. Meaning for every x value there can be a y value that is different than another y value for another x value.

Discrete Math: Combinatorics and recursion
3. Let S be a set of size 37, and let x, y, and z be three distinct elements of S. How many subsets of S are there that contain x and y, but do not contain z?
(a) $2^{33}$
(b) $2^{34}$
(c) $2^{35}$
(d) $2^{37}-2^{35}-2^{36}$
(d) none of the above
Why is it B) I thought there is size 37 so it is 37 - 2 Is it because there is size 37 and for x and y; you do 37-2. but you cannot have z so you minus another 1. so $37-<!--\; -\; -->2-<!--\; -\; -->1=34$; $2^{34}$
12. The Fibonacci numbers are defined as follows: $f0=0,f1=1$, and $fn=fn-1+fn-2$ for $n\ge <!--\; \ge \; -->2$. Consider again the recursive algorithm Fib, which takes as input an integer $n\ge <!--\; \ge \; -->0$:
Algorithm Fib(n):
if $n=0orn=1$
then $f=n$
else $f=Fib(n-1)+Fib(n-2)$
end if;
return f
For $n\ge <!--\; \ge \; -->3$, run algorithm Fib(n) and let an be the number of times that Fib(2) is called. Which of the following is true?
(a) For $n\ge <!--\; \ge \; -->3$, $a_n=f_{n-1}$
(b) For $n\ge <!--\; \ge \; -->3$, $a_n=f_n$
(c) For $n\ge <!--\; \ge \; -->3$, $a_n=f_{n+1}$
(d) For $n\ge <!--\; \ge \; -->3$, $a_n=-1+f_n$
So if it is $$n=3$$ i will call fib(2) 1 time and if $n=4$ then fib(2) is called 2 times
How do I put this into an equation like above?

Let G be a 3-regular plane graph with 12 faces. How many vertices does G have?
This would be pretty easy to solve if I knew that G is connected by using Eulers formula $|V|-<!--\; -\; -->|E|+|F|=2$.
But I don't know how to show that G is connected. Am I on the wrong path? Or is there some combinatorial argument to count the vertices?

Coefficient of expansion discrete math
What is the coefficient of $x^{12}{y}^{12}$ in the expansion of $(3x-<!--\; -\; -->7y{)}^{24}$? I am just checking if I answered it in the correct way. Since its the expansion to the power of 24 and $(xy{)}^{12}=(x^{12})(y^{12})$ then i Just substituted 3x and -7y with x and y. I got $(3x{)}^{12}(-<!--\; -\; -->7y{)}^{12}(\genfrac{}{}{0ex}{}{24}{12})$. Is my reasoning correct or is there much more to this.

Let $h=g\circ <!--\; \circ \; -->f\circ <!--\; \circ \; -->g$ where $f:<!--\; :\; -->\mathbb{R}\to <!--\; \to \; -->\mathbb{Z}$ is the floor function and $g:<!--\; :\; -->\mathbb{R}\to <!--\; \to \; -->\mathbb{R}:<!--\; :\; -->x\mapsto <!--\; \mapsto \; -->-<!--\; -\; -->x$.
(i) Compute h(3.4), h(7) and h(-1.3).
(ii) Describe what h is doing to a general real number x.

I wanted to ask here if this proving process were correct:
$f(a)=$ a div d
We must show this is an onto function
If it is onto, then $\mathrm{\forall <!--\; \forall \; -->}y$ in the codomain, $\mathrm{\exists <!--\; \exists \; -->}a$ in the domain such that $f(a)=y$
Consider an arbitrary y in the codomain. We know that $f(a)=y$ if and only if a div d $=y$.
a div $d=y$ implies that $a=dy+r$ .
But then $f(dy+r)=y$ because:
$f(dy+r)$= $(dy+r)$ div d meanining that
$dy+r=dq+k$ where q=$(dy+r)$ div d
Now since $a=dy+r$, a mod d is equal to $(dy+r)$ mod d so $k=r$
This means that:
$dy+r=dq+r$, so the quotient q of $dy+r$ is y. But then $f(dy+r)=y$, so for an arbitray y, there is an element $dy+r$ in the domain such that $f(dy+r)$ is y, as we wanted to show.

Relations between ordered pairs.
I am completely confused about this question, everytime I look back onto it I have a different idea on how to interpret it. Any help is appreciated.
A relation R is defined on ${\mathbb{Q}}^2$ by (a,b)R(c,d) if and only if there exists a real number $x\ge <!--\; \ge \; -->1$ such that $a=dx$ and $c=bx$.
I need to show what type of relation this is, e.g. is it reflexive, transitive, symmetric....? Right now, I am just having a lot of trouble on how to interpret this and how to actually come up with a way of proving this.

Discrete Math (Combination with repetition)
The employee is distributing 7 objects among 4 containrs (Xi's).
Assuming the containers are $X1+X2+X3+X4=7$ where $Xi>=0\mathrm{\forall <!--\; \forall \; -->}1<=i<=4$.
Determine all integer solutions.

Binary strings and discrete math
Let S be the set of binary strings of length 30 with 10 1’s and 20 0’s. Let A be the set of the first 30 positive integers {1,2,3,…,30}. Let B be the set of all subsets of A containing 10 numbers. Find a one-to-one correspondence between S and B.

Find Upper Bound for $T(n)=T(n-<!--\; -\; -->1)+T(\frac{n}{2})+n$ with recursive tree method.

How to choose witnesses for asymptotic growth?
Prove:
$$4n^5–50n^2+10n\in <!--\; \in \; -->\mathrm{\Theta <!--\; \Theta \; -->}(n^5)$$
$$0\le <!--\; \le \; -->c_1(n^5)\le <!--\; \le \; -->4n^5–50n^2+10n\le <!--\; \le \; -->c_2(n^5)$$
$$0\le <!--\; \le \; -->c_1\le <!--\; \le \; -->4–50/n^3+10/n^4\le <!--\; \le \; -->c_2$$
Now, I know for the middle portion, so long as $n\ge <!--\; \ge \; -->3$ then it will approach 4, leaving me with
$$0\le <!--\; \le \; -->c_1\le <!--\; \le \; -->4\le <!--\; \le \; -->c_2$$
However, I don't understand how to get my exact witnesses. Everything I found on the internet, people choose their witnesses but don't really explain how. Can I just arbitrarily choose anything between 0 and 4 for $c_1$ and anything above 4 for $c_2$?