Explore Discrete Math Examples

Bounding $A(n,d)=max\{M|$ exists a code with parameters n,M,d}
I would like to prove that this lower bound of $A(n,d)=max\{M|$ exists a code with parameters n,M,d} (where n is the length of the block code, M the number of words of the code, and d, the minimal distance of the code), holds:
$\frac{2^n}{\underset{i=0}{\overset{d-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n}{i})}\le <!--\; \le \; -->A(n,d)$
For trying to see this, I have tried to connect this inequality with the cardinal of the ball of radius $d-<!--\; -\; -->1$, that is $\underset{i=0}{\overset{d-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n}{i})2^i$, so for sure, that quantity is less than $2^n\underset{i=0}{\overset{d-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n}{i})$. But I don't see if this is or not helping me at all... I would appreciate some guidance, help, hint,... Thanks!

Closed form for a sequence of a language variables
We developed a new language based on some rules: The language limits all variable names to zero or more letters from the set {b, e, d} followed by zero or more digits from the set {0, 1, 2, 3}. In other words, the empty string, b, 12, bed100, and bedb2012 are all legal variable names. However, 1b and be2d are not valid variable names. There is exactly 1 legal variable name of length 0, namely the empty string. There are exactly 7 legal variable names of length 1: b, e, d, 0, 1, 2, and 3. There are exactly 37 legal variable names of length 2: bb, be, bd, b0, b1, b2, b3, eb, ee, ed, e0, e1, e2, e3, db, de, dd, d0, d1, d2, d3, 00, 01, 02, 03, 10, 11, 12, 13, 20, 21, 22, 23, 30, 31, 32, and 33. Your task is to find a closed-form expression for the number of legal variable names of length n in the language.
I have tried finding the sequence first so that I can find the closed-form but I am stuck in the first part, can't even go further. Like, the difference between length 0,1,2 is 1,7,37. Every form I take, I can't find the difference d from it. Not only that, how am I supposed to find a closed-form for it?

Discrete math: determining whether a relation is a function
In my book, they ask us to determine whether the following is a function:
$S\subseteq <!--\; \subseteq \; -->\mathbb{R}\times <!--\; \times \; -->\mathbb{R}$ with $S=\{(a,b)\mid <!--\; \mid \; -->a+b=10\}$
For previous problems like this, they gave us a relation of ordered pairs on a set, which I easily plugged into domain/codomain and mapped to each other to determine if it satisfied the definition of a function. For this question, I have no idea where to start.
I know it is saying that S is a subset of $\mathbb{R}\times <!--\; \times \; -->\mathbb{R}$ (not sure what $\mathbb{R}\times <!--\; \times \; -->\mathbb{R}$ means here), and that $S=$ a set where $\{(a,b)\text{such that}a+b=10\}$. I'm not sure how to use this information.

Prove that set is uncountable
$\{f\mid <!--\; \mid \; -->f:\mathbb{N}\to \{0,1\}\}-<!--\; -\; -->\{f\mid <!--\; \mid \; -->f:\{0,1\}\to \mathbb{N}\}$
Where f is a function. Prove that this difference is an uncountable set.
I am pretty stuck on how to start this problem. I understand that the set from 0 to 1 is uncountable while the set of natural numbers is countable, but I can't seem to start off the proof.

In how many ways can we distribute 2 types of gifts?
The problem: In how many ways can we distribute 2 types of gifts, m of the first kind and n of the second to k kids, if there can be kids with no gifts?
From the stars and bars method i know that you can distribute m objects to k boxes in $(\genfrac{}{}{0ex}{}{m+k-<!--\; -\; -->1}{k-<!--\; -\; -->1})$ ways. So in my case i can distribute m gifts to k kids in $(\genfrac{}{}{0ex}{}{m+k-<!--\; -\; -->1}{k-<!--\; -\; -->1})$ ways, same for n gifts i can distribute them in $(\genfrac{}{}{0ex}{}{n+k-<!--\; -\; -->1}{k-<!--\; -\; -->1})$ ways. So now if we have to distribute m and n gifts we can first distribute m gifts in $(\genfrac{}{}{0ex}{}{m+k-<!--\; -\; -->1}{k-<!--\; -\; -->1})$ ways, then n gifts in $(\genfrac{}{}{0ex}{}{n+k-<!--\; -\; -->1}{k-<!--\; -\; -->1})$ ways, so in total we have:
$(\genfrac{}{}{0ex}{}{m+k-<!--\; -\; -->1}{k-<!--\; -\; -->1})\cdot <!--\; \cdot \; -->(\genfrac{}{}{0ex}{}{n+k-<!--\; -\; -->1}{k-<!--\; -\; -->1})\text{ways.}$.
Is my reasoning correct?
What about when we have to give at least 1 gift to each kid, can we do that in
$(\genfrac{}{}{0ex}{}{m-<!--\; -\; -->1}{k-<!--\; -\; -->1})\cdot <!--\; \cdot \; -->(\genfrac{}{}{0ex}{}{n+k-<!--\; -\; -->1}{k-<!--\; -\; -->1})+(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->1}{k-<!--\; -\; -->1})\cdot <!--\; \cdot \; -->(\genfrac{}{}{0ex}{}{m+k-<!--\; -\; -->1}{k-<!--\; -\; -->1})\text{ways?}$

How to approach this discrete graph question about Trees.
A tree contains exactly one vertex of degree d, for each $d\in <!--\; \in \; -->\{3,9,10,11,12\}$.
Every other vertex has degrees 1 and 2. How many vertices have degree 1?
I've only tried manually drawing this tree and trying to figure it out that way, however this makes the drawing far too big to complete , I'm sure there are more efficient methods of finding the solution.
Could someone please point me in the right direction!

Representing a sentence with quantified statements
My approach to this question: $\mathrm{\exists <!--\; \exists \; -->}x(P(x)\to <!--\; \to \; -->R(x))$
I cannot verify if my answer is correct, any help to verify my answer would be appreciated and if I did wrong any help to explain why would also be appreciated.

Exercise involving DFT
The fourier matrix is a transformation matrix where each component is defined as $F_{ab}={\mathrm{\omega <!--\; \omega \; -->}}^{ab}$ where $\mathrm{\omega <!--\; \omega \; -->}=e^{2\mathrm{\pi <!--\; \pi \; -->}i/n}$. The indices of the matrix range from 0 to $n-<!--\; -\; -->1$ (i.e. $a,b\in <!--\; \in \; -->\{0,...,n-<!--\; -\; -->1\}$)
As such we can write the Fourier transform of a complex vector v as $\stackrel{^<!--\; ^\; -->}{v}=Fv$, which means that
${\stackrel{^<!--\; ^\; -->}{v}}_f=\underset{a\in <!--\; \in \; -->\{0,...,n-<!--\; -\; -->1\}}{\sum <!--\; \sum \; -->}{\mathrm{\omega <!--\; \omega \; -->}}^{af}{v}_a$
Assume that n is a power of 2. I need to prove that for all odd $c\in <!--\; \in \; -->\{0,...,n-<!--\; -\; -->1\}$, every $d\in <!--\; \in \; -->\{0,...,n-<!--\; -\; -->1\}$ and every complex vectors v, if $w_b=v_{cb+d}$, then for all $f\in <!--\; \in \; -->\{0,...,n-<!--\; -\; -->1\}$ it is the case that:
${\stackrel{^<!--\; ^\; -->}{w}}_{cf}={\mathrm{\omega <!--\; \omega \; -->}}^{-<!--\; -\; -->fd}\phantom{a}{\stackrel{^<!--\; ^\; -->}{v}}_f$
I was able to prove it for $n=2$ and $n=4$, so I tried an inductive approach. This doesn't seem to be the best way to go and I am stuck at the inductive step and I don't think I can go any further which indicates that this isn't the right approach.
Note that I am not looking for a full solution, just looking for a hint.

Combinatorial proof of $P_r^{n+1}=r!+r(P_{r-<!--\; -\; -->1}^n+P_{r-<!--\; -\; -->1}^{n-<!--\; -\; -->1}+\cdots <!--\; \cdots \; -->+P_{r-<!--\; -\; -->1}^r)$, where $P_r^n$ denotes the number of r- permutations of an n element set.

How to solve $\lfloor <!--\; \lfloor \; -->x\rfloor <!--\; \rfloor \; -->+\lfloor <!--\; \lfloor \; -->\frac{1}{x}\rfloor <!--\; \rfloor \; -->=1$?

Solution of linear congruence system
Solve $\{\begin{array}{l}2x\equiv <!--\; \equiv \; -->1(\mathrm{mod}5)\\ (5x+2)\equiv <!--\; \equiv \; -->2(\mathrm{mod}18)\end{array}$
Now I have read about the Chinese Remainder Theorem, which is particularily helpful in solving systems of linear congruence, but in this notice that the moduli are not relatively prime (pairwise). So, I cannot apply the theorem. In fact this leads me to a general question: If we are given as system:
$$\begin{gathered} a_{1}x\equiv <!--\; \equiv \; -->b_1(\mathrm{mod}{n}_1) \\ {a}_{2}x\equiv <!--\; \equiv \; -->b_2(\mathrm{mod}{n}_2) \\ {a}_{3}x\equiv <!--\; \equiv \; -->b_3(\mathrm{mod}{n}_3) \\ {a}_{4}x\equiv <!--\; \equiv \; -->b_4(\mathrm{mod}{n}_4) \\ ... \\ {a}_{r}x\equiv <!--\; \equiv \; -->b_r(\mathrm{mod}{n}_r) \end{gathered}$$
where the moduli are not pairwise relatively prime. How do we go about solving this system and what are the conditions that determine the solvability of the system.

Prove that a relation isn't transitive
Let $\begin{array}{r}{M}_R=\left(\begin{array}{ccc}1& 1& 0\\ 1& 1& 1\\ 0& 1& 0\end{array}\right)\end{array}$
Where $M_R$ is the relation matrix for a relation R. Is R reflexive, symmetric, antisymmetric or transitive?
I find that is Symmetric but isn't reflexive and antisymmetric, To verify if $M_R$ is transitive. I compute the Boolean product
$\begin{array}{r}\left(\begin{array}{ccc}1& 1& 0\\ 1& 1& 1\\ 0& 1& 0\end{array}\right)\odot <!--\; \odot \; -->\left(\begin{array}{ccc}1& 1& 0\\ 1& 1& 1\\ 0& 1& 0\end{array}\right)=\left(\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right)\end{array}$
That means that $M_R\odot <!--\; \odot \; -->M_R\ne <!--\; \ne \; -->M_R$, So $M_R$ isn't transitive. This is correct?

One box - one pair -2 different things to create 4 length
I am trying to solve exercise that has to do with Combinations and Permutations.I think it is Permutation because if it was Combination we care about the order. I have stacked in an exercise and I would like to give me your help.
How many with length 4 can I create from a box ( the box has 24 balls) with 1 pair at least that has successively white balls or successively black balls. The white balls are 7 the black balls are 17.
What I did is to take $P(n,r)=n!/(n-r)!$ so for Pwhite(7,2). I did this, because I take 7 black ball and I want 2 pair so I have the result $Pwhite(7,2)=7!(7-<!--\; -\; -->2)!=42$
Now the black Pblack(17,2) $P(17,1)=17!/(17-2)!$. I have the result 272
I want to do 4 length of 2white and 2black(as I understand - maybe I am wrong).

How do I find $9^{17}mod7$ without a calculator
I've seen some examples of similar problems but I don't understand how they were solved. How would I solve $9^{17}mod7$ without calculator.

Help with using the Schroeder-Bernstein Theorem?
Corresponding Counts [3 points]
Prove that $|\{x\in <!--\; \in \; -->\mathbb{R}|0\le <!--\; \le \; -->x\le <!--\; \le \; -->1\}|=|\{x\in <!--\; \in \; -->\mathbb{R}|4<x<7\}|$.

Equivalence of mathematical induction and strong induction
Mathematical induction and strong induction are equivalent. That is, each can be shown to be a valid proof technique assuming that the other is valid.
I interpret this to mean that neither proof technique can be used where the other could not be used. That being the case, why bother about strong induction at all?

You have {a,b,c} as your character set. You need to create words having 10 characters that must be chosen from the character set. Out of all the possible arrangements, how many words have exactly 6 as?

How to expand the following notation?
Can someone help me to expand the below equation for $N=2$ and $N=3$?
$f_i=\sum <!--\; \sum \; -->\{e_i:j\le <!--\; \le \; -->i\le <!--\; \le \; -->k\}$ for some $1\le <!--\; \le \; -->j\le <!--\; \le \; -->k\le <!--\; \le \; -->N$.
Here $e_i$ is the unit vector of coordinate i in an N-dimensional space.

The following cards are dealt with 3 people at random so that every one of them gets the same number of cards:
$R_1,R_2,R_3,B_1,B_2,B_3,Y_1,Y_2,Y_3$
where R, B and Y denote red, blue and yellow, respectively. Find the probability that everyone gets a red card.

How to inductively prove a graph property?
I'm stuck on Part 1, I can't find it inductively. The distance from $[k,k+1]$ is going to be a non-zero positive value with a vertex. Therefore, there is going to be a positive edge weight from w(uv) that goes from $[k,k+1].$. Therefore, there exists a v on the set of V such at $A[u,k+1]=A[v,k]+w(uv).$.
The Problem
Given an undirected graph $G=(V,E)$ with positive edge weights w(e) for each edge $e\in <!--\; \in \; -->E$, we want to find a dynamic programming algorithm to compute the longest path in G from a given source s that contains at most n edges.
To do this first define A[v,k] as the weight for the longest path from node s to node v of at most k edges.
1. First we need to prove an optimal sub-structure by induction. Show that if A[v,k] is the weight of the longest path, then for all $u\in <!--\; \in \; -->V$, there exists a $v\in <!--\; \in \; -->V$ such that $A[u,k+1]=A[v,k]+w(uv)$.
2. Describe a dynamic programming algorithm that finds the optimal length using part 1. Specifically: describe (1) the OPT recurrence (2) the running time of the iterative solution for computing the OPT table.