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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.

Throw 5 times a dice with 6 sides colored in: blue, red, yellow, green, orange (5 colors)
How many possibles answers for getting at least one cube with '3'?
Exactly one cube with '2', and exactly one with with '4'?

How to solve for the inverse of f for a floor function? I really need to learn the method.
I was doing some exercises on discrete maths and came across this question I don't know how to solve, nor can I find any relevant examples in my book.
Suppose $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ where $f(x)=\lfloor x/2\rfloor$, if $T=\{3,4,5\}$ , then what's the inverse of f(T)?
I know how to get f(T) but how to get its inverse? Could anyone teach me please?

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.

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?

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.
$$\begin{array}{|lll|}\hline p& q& pimp1\\ T& T& T\\ T& F& F\\ F& T& F\\ F& F& T\\ \hline\end{array}$$

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.

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 $c_2$?