Explore Discrete Math Examples

We are given the digits 1,2,4,5,6,7,9. How many even, 4-digit numbers, bigger than 5000, with distinct digits can we form?

Find the image of the function $f:[-10,10]\to R$ defined by $f(x)=x^2+2x+2$.
My attempt
I need to find the function image. To do this, I must replace x by the intervals of the function.
Intervals: [-10, 10]
x: $f(x)=x^2+2x+2$
So putting the two intervals, it would be:
$f(-<!--\; -\; -->10)=(-<!--\; -\; -->10{)}^2+2.(-<!--\; -\; -->10)+2$
$f(-<!--\; -\; -->10)=100+(-<!--\; -\; -->20)+2$
$f(-<!--\; -\; -->10)=100+(-<!--\; -\; -->18)$
$f(-<!--\; -\; -->10)=82$
$f(10)={10}^2+2.10+2$
$f(10)=100+20+2$
$f(10)=100+22$
$f(10)=122$
So the function image would be:
$Im=[82,122]$
Does it makes sense? I'm not entirely correct about the resolution, if anyone can look into it I'm grateful.

Is this true for chromatic numbers?
If we take the graph $G=(V(G),E(G))$ and partition the edges E(G) into k sets, forming k subgraphs of the form $H_i=(V(G),E(H_i))$ for $i\in <!--\; \in \; -->\{1,...,k\}$ such that $|E(H_i)|\ge <!--\; \ge \; -->1$ for each subgraph $H_i$. Furthermore, we let χ(G) be the chromatic number of the graph G.
Does it hold that $\underset{i=1}{\overset{k}{\prod <!--\; \prod \; -->}}\mathrm{\chi <!--\; \chi \; -->}(H_i)\ge <!--\; \ge \; -->\mathrm{\chi <!--\; \chi \; -->}(G)$? How would we prove it? I know that this holds when $k=2$. How about other values $k\ge <!--\; \ge \; -->3$?
For the case $k=2$ we can take $H,\stackrel{¯<!--\; ¯\; -->}{H}\subseteq <!--\; \subseteq \; -->G$ such that $H=(V(G),E(H))$ and $\stackrel{¯<!--\; ¯\; -->}{H}=(V(G),E\setminus <!--\; \setminus \; -->E(H))$. If the coloring $C_H:V(G)\to <!--\; \to \; -->[\mathrm{\chi <!--\; \chi \; -->}(H)]$ is a coloring of H and $C_{\stackrel{¯<!--\; ¯\; -->}{H}}:V(G)\to <!--\; \to \; -->[\mathrm{\chi <!--\; \chi \; -->}(\stackrel{¯<!--\; ¯\; -->}{H})]$ is a coloring of $\stackrel{¯<!--\; ¯\; -->}{H}$, we are able to construct a coloring $C_G$ of G with at most $\mathrm{\chi <!--\; \chi \; -->}(H)\cdot <!--\; \cdot \; -->\mathrm{\chi <!--\; \chi \; -->}(\stackrel{¯<!--\; ¯\; -->}{H})$ colors by letting $C_G(v)=(C_H(v),C_{\stackrel{¯<!--\; ¯\; -->}{H}}(v))$. We can also observe that every edge (u,v) in G belongs to either H or $\stackrel{¯<!--\; ¯\; -->}{H}$ and hence $C_G(u)$ differs from $C_G(v)$ in at least one of the coordinates.
Can this same argument potentially be generalized for the case of partitioning G into k subgraphs such that $|E(H_i)|\ge <!--\; \ge \; -->1$ for each subgraph?

Which one is reflexive, symmetric and transitive
I am doing an exercise where I need to find which one of the statements under are reflexive, symmetric and transitive. I will write what I think for each of the statements, could someone see if I am thinking right
Our domain is all real numbers:
a) $x+y=0$
This one is reflexive only for (0,0) but not for all real numbers, it is symmetric for (0,0) but not for all real numbers, but I dont know how to see if it is symmetrical.
b) $x=\pm <!--\; \pm \; -->y$. This is not reflexive for domain R, because (1,1) not true I dont know how to see if its not symmetrical or not.
c) $x=2y$. Not reflexive, (1,1) not true to find symmetric do we do it like this? $x=2y-<!--\; -\; -->>x/2=y$ means not symmetric dont know how to find transitive.
d) $xy\ge <!--\; \ge \; -->0$. Not reflexive, (1,1) not treu symmetric because xy, $y\cdot <!--\; \cdot \; -->x$ transitive?
e) $x\cdot <!--\; \cdot \; -->y=0$ same as d)
f) $x=1\text{or}y=1$. Not sure what to do

Number of arrangement of the word JANUARY such that N is before Y and no two vowels are next to eachother

How many ways we can load 100 indistinguishable boxes into 3 cars A,B,C,if every car can be loaded with 20 to 40 boxes?

How to recognize a geometric sequence while solving by unfolding?
I have a problem with how this is written as a geometric sequence. I got this screenshot from my math book. r should be 2 and a should be 1.
$S=\frac{1-<!--\; -\; -->2^{n-<!--\; -\; -->1}}{1-<!--\; -\; -->2}$
Step 2
How did they come up with n as the power of 2? And why is the sum being written backwards?
$\begin{array}{rl}{a}_n& =5+5(2)+5(2^2)+5(2^3)+\cdots <!--\; \cdots \; -->+5(2^{n-<!--\; -\; -->1})+2^n{a}_0\\ & =5(1+2+2^2+\cdots <!--\; \cdots \; -->+2^{n-<!--\; -\; -->1})+2^n(2)\\ & =5\frac{2^n-<!--\; -\; -->1}{2-<!--\; -\; -->1}+2(2^n)\end{array}$

In need of a combinatorial proof of $1\ast <!--\; \ast \; -->2+2\ast <!--\; \ast \; -->3+3\ast <!--\; \ast \; -->4+...+n(n+1)=2\ast <!--\; \ast \; -->(\genfrac{}{}{0ex}{}{n+2}{3})$ for natural numbers n.
For my discrete maths class, we are learning combinatorial proofs, and my prof handed this problem as practice. I'm aware of the process of a combinatorial proof (counting the same thing in two different ways), but I have no idea how to start this problem or how to relate the LHS to a counting problem.
All I know is the RHS is twice the number of ways to make 3 element subsets of a set of $n+2$ elements. A possible hint I found from a related problem in my book is to consider a set $S=\{0,1,2,3,...,n,n+1\}$ which has $n+2$ elements.

Which connective to use when translating the statement $\forall x(-2<x<3)$ to words?
Does the above read as “for all x, $x>-<!--\; -\; -->2$ and $<3$" or as "$x>-<!--\; -\; -->2$ or $<3$"?
I want to know which one to use, in order to be able to determine how to correctly negate the statement.

Show that $\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}kn=\mathrm{O}(n^3)$

Justification for g being a one to one function
If f and $f\circ <!--\; \circ \; -->g$ are one to one functions, is g also a one to one function? Justify your answer.
I am guessing that g should be a one to one function.
So a function F is injective if:
$\mathrm{\forall <!--\; \forall \; -->}a\mathrm{\forall <!--\; \forall \; -->}b(F(a)=F(b)\Rightarrow <!--\; \Rightarrow \; -->a=b)$
So since f is injective:
$\mathrm{\forall <!--\; \forall \; -->}a\mathrm{\forall <!--\; \forall \; -->}b(f(a)=f(b)\Rightarrow <!--\; \Rightarrow \; -->a=b)$
And $f\circ <!--\; \circ \; -->g$ is also injective:
$\mathrm{\forall <!--\; \forall \; -->}a\mathrm{\forall <!--\; \forall \; -->}b(f\circ <!--\; \circ \; -->g(a)=f\circ <!--\; \circ \; -->g(b)\Rightarrow <!--\; \Rightarrow \; -->g(a)=g(b))$
$\mathrm{\forall <!--\; \forall \; -->}a\mathrm{\forall <!--\; \forall \; -->}b(g(a)=g(b)\Rightarrow <!--\; \Rightarrow \; -->a=b)$
Does this show that g must also be injective?

Poker and Combinatorics (Don't Mix): How to solve this problem?
Does anyone know the solution to this problem? I'm not sure if it's asking to find the cardinality first, and whether I should use the additive counting principle or the multiplicative one? Would help me a lot, thanks!
Given is a standard 52-card poker game. A hand is a non-ordered selection of five Cards from the game. Consider the following question and the answer to it:
Question: How many hands are there that have all four colors?
Answer: We choose a card from each color, so there are $({13}^4)$ possibilities for this. 48 possibilities remain when we choose the fifth card. So there are a total of $({13}^4)\times <!--\; \times \; -->48$ hands that contain all four suits.
Is the solution correct? If not, what is wrong with the answer and what is the correct solution?

Probability / Combinatoric of fair roulette question
In a fair roulette there are 4 sections, numbered 1 to 4.
A player spins the roulette's pointer 10 times independently.
What is the chance that the pointer stops at each one of the 4 sections at least one?

How to write $(2n-<!--\; -\; -->1)(2n-<!--\; -\; -->2)(2n-<!--\; -\; -->3)(2n-<!--\; -\; -->4)$ terms of n factorials.
I want to get the final equations in terms of n, for example $n=3$, then the equation is:
$(2n-<!--\; -\; -->1)\ast <!--\; \ast \; -->(2n-<!--\; -\; -->2)\ast <!--\; \ast \; -->(2n-<!--\; -\; -->3)$, then the equation is: $(2n-<!--\; -\; -->1)(2n-<!--\; -\; -->2)(2n-<!--\; -\; -->3)(2n-<!--\; -\; -->4)$.

Find the number of all sequences $\{a_n\}$ in $\{-<!--\; -\; -->5,-<!--\; -\; -->4,-<!--\; -\; -->3,...,0,1,...,100\}$ such that $|a_n|<|a_{n+1}|$

Prove that The sum of the cube of the first n odd numbers $n^2(2n^2-<!--\; -\; -->1)$.
That is
$\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}(2i-<!--\; -\; -->1{)}^3=n^2(2n^2-<!--\; -\; -->1)$
First I tried to solve it by induction:
For $n=1$ we have
$\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}(2i-<!--\; -\; -->1{)}^3=1$
and $n^2(2n^2-<!--\; -\; -->1)=1$
Let $n\in <!--\; \in \; -->\mathbb{N},$,
$Assume\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}(2i-<!--\; -\; -->1{)}^3=n^2(2n^2-<!--\; -\; -->1)$
$Prove\underset{i=1}{\overset{n+1}{\sum <!--\; \sum \; -->}}(2i-<!--\; -\; -->1{)}^3=(n+1{)}^2(2(n+1{)}^2-<!--\; -\; -->1)$
I know that $\underset{i=1}{\overset{n+1}{\sum <!--\; \sum \; -->}}(2i-<!--\; -\; -->1{)}^3=\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}(2i-<!--\; -\; -->1{)}^3+(2(n+1)-<!--\; -\; -->1{)}^3$
But i was not able to reach a solution. If anyone can give a small hint, that would be helpful.

What to do if the Extended Euclidean Algorithm terminates in one step?
I am trying to solve the linear congruence
$14x\equiv <!--\; \equiv \; -->1(\mathrm{mod}113)$
So I first find $gcd(14,113)=1$. However this means that:
$113=14(8)+1$
There is only one step needed. If I don't have any other equations, how do I find the inverse? Thank you for any help!

I wish to solve exactly this formula involving sums and products
$$\begin{gathered} ⟨n_l⟩={[1+\underset{k\ne <!--\; \ne \; -->l}{\sum <!--\; \sum \; -->}\left(e^{bN(l-<!--\; -\; -->k)}\frac{\underset{j\ne <!--\; \ne \; -->l}{\prod <!--\; \prod \; -->}(1-<!--\; -\; -->e^{b(l-<!--\; -\; -->j)})}{\underset{j\ne <!--\; \ne \; -->k}{\prod <!--\; \prod \; -->}(1-<!--\; -\; -->e^{b(k-<!--\; -\; -->j)})}\right)]}^{-<!--\; -\; -->1}(N+\underset{h\ne <!--\; \ne \; -->l}{\sum <!--\; \sum \; -->}\frac{1}{1-<!--\; -\; -->e^{b(h-<!--\; -\; -->l)}})-<!--\; -\; --> \\ \underset{h\ne <!--\; \ne \; -->l}{\sum <!--\; \sum \; -->}\left\{{[1+\underset{k\ne <!--\; \ne \; -->h}{\sum <!--\; \sum \; -->}\left(e^{bN(h-<!--\; -\; -->k)}\frac{\underset{j\ne <!--\; \ne \; -->h}{\prod <!--\; \prod \; -->}(1-<!--\; -\; -->e^{b(h-<!--\; -\; -->j)})}{\underset{j\ne <!--\; \ne \; -->k}{\prod <!--\; \prod \; -->}(1-<!--\; -\; -->e^{b(k-<!--\; -\; -->j)})}\right)]}^{-<!--\; -\; -->1}\left(\frac{1}{1-<!--\; -\; -->e^{b(l-<!--\; -\; -->h)}}\right)\right\} \end{gathered}$$
where b is a positive real number, N is a natural number, and all the sums and products are understood to run from 0 to $M-<!--\; -\; -->1$, where M is yet another natural number (all the l, h, j, k indices are therefore bound to this interval, with the appropriate restrictions written under the sums and the products). Eventually I would also like to take the limit as M goes to infinity, but I'm sure how well I can do that.
Do you think there is a hope to massage this formula to make it a little less ugly? For the moment, I've been trying to solve it numerically...

How to prove $p(x)\ge <!--\; \ge \; -->m^2/4$
Let $\mathcal{F}$ be a family of m subsets of a finite set X. For $x\in <!--\; \in \; -->X$, let p(x) be the number of pairs (A,B) of sets $A,B\in <!--\; \in \; -->\mathcal{F}$ such that either $x\in <!--\; \in \; -->A\cap <!--\; \cap \; -->B$ or $x\notin A\cup <!--\; \cup \; -->B$. Prove that $p(x)\ge <!--\; \ge \; -->m^2/2$.
In the book I'm studying, writer has written the following hint:
Hint: Let d(x) be the degree of x in $\mathcal{F}$, and observe that $p(x)=d(x{)}^2+(m-<!--\; -\; -->d(x){)}^2$.
I was wondering if someone could help me about my problem. Thanks in advance.

Is this predicate-logic formula a tautology?
$F(x)\to <!--\; \to \; -->\mathrm{\forall <!--\; \forall \; -->}xF(x)$
My teacher says that the formula is a tautology. Because when the antecedent is false, all the formula should always true, and when the antecedent is true, consequent can be infered by using $\mathrm{\forall <!--\; \forall \; -->}+$ rule.