Explore Discrete Math Examples

Let R be a relation on E. Demonstrate that:
- R is reflexive if and only if $I{d}_E\subseteq <!--\; \subseteq \; -->R$;
- R is symmetric only and only if $R=R-<!--\; -\; -->1$.

Proofs in Discrete Math
$\mathrm{\forall <!--\; \forall \; -->}n\in <!--\; \in \; -->N+$, n composite $\to <!--\; \to \; -->\mathrm{\exists <!--\; \exists \; -->}p\in <!--\; \in \; -->N+$, p is prime and $p\le <!--\; \le \; -->\sqrt{n}$ and p|n.
Am I supposed to prove that $p\le <!--\; \le \; -->\sqrt{n}$ and p|n when n is composite and p is prime? Could someone fix my translation if I'm wrong?

Rewriting biconditional statements into conditional statement without arrows (discrete math)
I'm stuck on rewriting biconditional statements to conditional statements without using arrows. the original is as follows:
$((P\to <!--\; \to \; -->(\sim <!--\; \sim \; -->Q))\leftrightarrow <!--\; \leftrightarrow \; -->r)$ and what i (believe) the correct answer is:
$((\sim <!--\; \sim \; -->PV\sim <!--\; \sim \; -->Q)\leftrightarrow <!--\; \leftrightarrow \; -->r)$
$[\sim <!--\; \sim \; -->(\sim <!--\; \sim \; -->PV\sim <!--\; \sim \; -->Q)Vr]\Lambda [(\sim <!--\; \sim \; -->rV(\sim <!--\; \sim \; -->PV\sim <!--\; \sim \; -->Q)]$If anyone would like to correct and explain why i'm wrong or verify it's correct I would greatly appreciate it.

How to determine class complexities of a discrete math problem?
I am having trouble understanding how to determine the class complexities (NP complete, NP hard, P,...) of a discrete math problem. Could someone show the detailed procedure with explanations on these three problems? Assume that $P\ne <!--\; \ne \; -->NP.$ Please give math strict explanations.
1. Is it in electric circuit that connects each of n cities, the shortest length of a transmission line that passes only once through each city and connects all cities in a closed network is less than k?
2. Given a sequence of n points of two dimensional coordinate system and indices of two points from that sequence. Is the shortest closed path that passes through all points shorter from the shortest path between given two points?
3. In a social network of n people, does the largest set of people that knows each other contains more than k people?

Discrete Math-Computing Summations
So I'm asked to compute a summation with an upper limit $k=20$ and lower limit $k=1$, where:
$B_k=0$ when $k=1$, and $B_k={\displaystyle \frac{1}{(k^2-<!--\; -\; -->1)}}$, for $k>1$.

Probability - discrete math
In a particular group of people, 10 people are right handed and 4 are left handed. If 5 of these people are chosen at random, What is the probability that exactly 1 left handed person is selected?

Discrete Math Proof
This is one of my midterm review questions. I am not sure where to start. Should I use induction to prove that, if so, how should I do it?
Recall that $\mathbb{N}=\{1,2,3,4,\dots <!--\; \dots \; -->\}$. Prove that $\mathbb{N}$ and $B=\mathbb{N}\cup <!--\; \cup \; -->\{-<!--\; -\; -->1,0\}$ have the same cardinality.

Writing statements into symbols Discrete Math.
The variable x represents students, F(x) means "x is a freshman", and M(x) means "x is a math major"
a) some freshme are math majors? $\mathrm{\exists <!--\; \exists \; -->}x:F(x)⟹<!--\; ⟹\; -->M(x)$
b) Every math major is a freshman? $\mathrm{\forall <!--\; \forall \; -->}x:M(x)⟹<!--\; ⟹\; -->M(x)$
c) No math major is a freshman? $\mathrm{\neg <!--\; \neg \; -->}\mathrm{\forall <!--\; \forall \; -->}x:M(x)⟹<!--\; ⟹\; -->F(x)$

For a problem on a discrete math assignment, I am asked to find which of the following statements is true, but I am unsure if I'm interpreting it correctly as I don't know how to interpret every symbol yet. Here are the statements:
$\mathrm{\exists <!--\; \exists \; -->}!x\in <!--\; \in \; -->Z,\mathrm{\forall <!--\; \forall \; -->}y\in <!--\; \in \; -->Z,xy=x.$
$\mathrm{\exists <!--\; \exists \; -->}!x\in <!--\; \in \; -->Z,\mathrm{\forall <!--\; \forall \; -->}y\in <!--\; \in \; -->Z,xy=y$
Should I interpret this as "there exists exactly one x for all y such that ..." or should I interpret this as "there exists exactly one x for each y such that ..."

Let $S=\{a,b\}$ be a two-element set. How many functions $f:S\to <!--\; \to \; -->S$ have the property $f\circ <!--\; \circ \; -->f$ is the identity function?
one or two or three?
This is my university discrete math question, however, i am really out of thought for this one. I understand the identity function is that we give $f(x)=x$ and $g(x)=x$, but for $f\circ <!--\; \circ \; -->f$, the only thing i can thought is $f(x)\times <!--\; \times \; -->f(x)$, am i right?

Discrete math - Prove that a tree with n nodes must have exactly $n-<!--\; -\; -->1$ edges? I have researched the solution but I haven't founded yet. I know of course, a tree with n nodes must have exactly $n-<!--\; -\; -->1$ edges. But, I can't prove it.

I need help with one of those question where you have to count the number of ways you can place r objects into n distinct boxes kinds. I was wondering if someone could solve an example in detail.
How many positive integers greater than equal to 1 and less than or equal to 10010 have the property that the sum of their digits is exactly 9?

Discrete Math Proof: $A\cup <!--\; \cup \; -->B$
The problem I'm having trouble proving is the following:
$A\cup <!--\; \cup \; -->B=(A\cap <!--\; \cap \; -->BC)\cup <!--\; \cup \; -->(AC\cap <!--\; \cap \; -->B)\cup <!--\; \cup \; -->(A\cap <!--\; \cap \; -->B)$, where C denotes complement of a set

My discrete maths book Gives proof for The sum of geometric series as:
$S=\underset{j=0}{\overset{n}{\sum <!--\; \sum \; -->}}a{r}^j=\frac{(a{r}^{n+1}-<!--\; -\; -->a)}{r-<!--\; -\; -->1}$
When $r!=1$,
And the next line reads:
Clearly, when $r=1$, then, the sum is:
$(n+1)a$
I don't understand the clearly part, What I could clearly see was only then when $r=1$, the expression took an indeterminate form.
How was the $(n+1)a$ expression obtained?

"Prove the following used the method of contradiction: The sum of two consecutive integers is always odd."
I thought this proof would be a straightforward direct proof. So, the contradiction would be "The sum of two integers is always even." Sparing the rigorous details: an integer n added to another integer $(n+1)$ leads to $(2n+1)$, which contradicts the statement, since $2n+1$ is the representation of an odd number.
My teacher, however, proved this with two cases. The first case: a direct proof, using my strategy above, for $n+(n+1)$. The second case, basically a similar proof to the one in the first case but now using $(n-<!--\; -\; -->1)+n$. This second case is what has confused me. Isn't this step a bit redundant? Is it necessary? Does it enhance the proof, or just add superfluous information to it?

How can I prove this equation for my discrete math project?
Prove:
${\displaystyle \underset{i\ge <!--\; \ge \; -->1}{\prod <!--\; \prod \; -->}\frac{1}{1-<!--\; -\; -->x{q}^i}}={\displaystyle \underset{k\ge <!--\; \ge \; -->0}{\sum <!--\; \sum \; -->}\frac{x^k{q}^{k^2}}{(1-<!--\; -\; -->q)(1-<!--\; -\; -->q^2)\cdots <!--\; \cdots \; -->(1-<!--\; -\; -->q^k)(1-<!--\; -\; -->xq)(1-<!--\; -\; -->x{q}^2)\cdots <!--\; \cdots \; -->(1-<!--\; -\; -->x{q}^k)}}$
One of the hints that was given was Let O be the collection of integer partitions with only parts of odd size. Then both sides are the generating function $\underset{p\mathrm{ϵ<!--\; ϵ\; -->}O}{\sum <!--\; \sum \; -->}{x}^{l(p)}{q}^{|p|}$ where l(p) is the number of parts of p and |p| is the sum of the parts.

Discrete math: proofs
For any three integers x, y, and z, if 𝒚 is divisible by x and z is divisible by y, then z is divisible by x.
definition: An integer n is divisible by an integer d with $D\ne <!--\; \ne \; -->0$, denoted d|n, if and only if there exists an integer k such that $n=dk$.
Can anyone help, with this problem I don't know how to approach it should I use proof by cases where every number is odd or even. Should I use direct proof or indirect proof?

Suppose you are trying to get rid of your leftover Halloween chocolate. You decide to make individualized gift bags to give to your lecturers. You have 4 different types of chocolate to choose from. How many unique gift bags can you create with 10 items per bag such that each bag has at least one of each type of chocolate.

Prove that among any 100 integers there are always two whose difference is divisible by 99. How can I prove this?

Prove or disprove (discrete math)
Prove or disprove the following statement. The difference of the square of any two consecutive integers is odd
This is working step: let $m,m+1$ be 2 consective integers:
$(m+1{)}^2-<!--\; -\; -->m^2$
$m^2+1+2m-<!--\; -\; -->m^2$
$1+2m$
If m is odd then $2m=\text{even}$, if m is even then $2m=\text{even}$, then adding 1 will make it odd.
Can you please advise me if my working is the right step and could I answer like this?