Explore Discrete Math Examples

Combinatorics of sports outcomes
I am trying to figure out how many possible combinations of 6 fighters can be made out of a pool of 22 fighters. These are 1v1 fights (11 fights) so I don’t want any 2 fighters fighting each other to be in the same combination. I imagine it has to be something similar to 22 choose 6 but that wouldn’t account for the fighters against each other. Order does not matter.

Rewriting a logical statement with multiple quantifiers
I am currently reading "Discrete Mathematics with Applications" by Susanna Epp. Not all questions have answers in this book. I stumbled upon question 11 from Section 3.3
Let S be the set of students at your school, let M be the set of movies that have ever been released, and let V(s,m) be "student s have seen movie m." Rewrite the following statement without the symbol $\mathrm{\forall <!--\; \forall \; -->}$, the symbol $\mathrm{\exists <!--\; \exists \; -->}$ or variables.
$\mathrm{\exists <!--\; \exists \; -->}s\in <!--\; \in \; -->S$ and $\mathrm{\exists <!--\; \exists \; -->}t\in <!--\; \in \; -->S$ such that $s\ne <!--\; \ne \; -->t$ and $\mathrm{\forall <!--\; \forall \; -->}m\in <!--\; \in \; -->M,V(s,m)\to <!--\; \backslash rightarrow\; -->V(t,m)$
I interpreted this as "There are two students who watched all released movies". Is it right? If yes why do we need an if-then operator? If not what is the correct way?

Generating Function of Riordan numbers
I would like to find generating function of f(n), where f(n) is defined as following: $f(n)=\underset{k}{\overset{n}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n}{k})(-<!--\; -\; -->1{)}^{n-<!--\; -\; -->k}{C}_k\text{.}$
With $C_k=\frac{1}{k+1}(\genfrac{}{}{0ex}{}{2k}{k})$ ($C_k$ is the $k^{th}$ Catalan's number).

Find the number of solutions in integers to $n_1+n_2+n_3+n_4=12$ satisfying $0\le <!--\; \le \; -->n_1\le <!--\; \le \; -->4,0\le <!--\; \le \; -->n_2\le <!--\; \le \; -->5,0\le <!--\; \le \; -->n_3\le <!--\; \le \; -->8,\text{and }0\le <!--\; \le \; -->n_4\le <!--\; \le \; -->9$.

I want to make sure that this is right with respect to the cardinality
Consider the following sets: $A=\{0,1\}$, $B=\{\{0,1\}\}$ and $C=A\cup <!--\; \cup \; -->B$. Enumerate the following sets and report their cardinality:
No. 1: $2^C$
No. 2: $C\times <!--\; \times \; -->C$ (cross-product)
No. 1: $\{\},\{0\},\{0,1\},\{1\},\{0,0\},\{0,1\},\{1,0\},\{1,1\}$, so we have 8, so $2^8$ would be 256 and that would be the cardinality.
No. 2: $\{0\},\{0,1\},\{1\},\{0,0\},\{0,1\},\{1,0\},\{1,1\},\{0\},\{0,1\},\{1\},\{0,0\},\{0,1\},\{1,0\},\{1,1\}$ so it would be 14 for the cardinality.
I just want to make sure if these are right, and if not, find out where I went wrong.

Biconditionals and Conjunctions in Truth Tables
Given that a biconditional $p⟺<!--\; ⟺\; -->q$ is True what can be concluded from the statement $\mathrm{\neg <!--\; \neg \; -->}p\wedge <!--\; \wedge \; -->\mathrm{\neg <!--\; \neg \; -->}q$?
In a worded example:
I wear my running shoes if and only if I exercise. (True)
I am not exercising AND I am not wearing my running shoes. (?)
If we set up a truth table, the biconditional is True in two of the four occurrences, but we see that $\mathrm{\neg <!--\; \neg \; -->}p\wedge <!--\; \wedge \; -->\mathrm{\neg <!--\; \neg \; -->}q$ is both True and False, which would mean there is no conclusion, correct?

Intuition behind multiplying (or composing) permutations.
I'm trying to grasp the intuition for permutations and their multiplication. So far this has been my intuitive understanding: A permutation is merely a shuffling of the symbols. Take for example $\mathrm{\sigma <!--\; \sigma \; -->},\mathrm{\pi <!--\; \pi \; -->}\in <!--\; \in \; -->S_4$ given by, $\mathrm{\sigma <!--\; \sigma \; -->}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 3& 2& 1& 4\end{array}\right)$ and $\mathrm{\pi <!--\; \pi \; -->}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 4& 1& 3\end{array}\right)$
I could rewrite them as a 4-tuple: $\mathrm{\sigma <!--\; \sigma \; -->}=(3,2,1,4)$ and $\mathrm{\pi <!--\; \pi \; -->}=(2,4,1,3)$ as permutations of {1,2,3,4} and so $\begin{array}{}\text{(\#)}& \mathrm{\pi <!--\; \pi \; -->}\circ <!--\; \circ \; -->\mathrm{\sigma <!--\; \sigma \; -->}=(2,4,1,3)\circ <!--\; \circ \; -->(3,2,1,4)=(1,4,2,3)\end{array}$
I understand how to get the result. I know how to multiply (or compose) two permutations.
My Question: What happened in equation # and what's going on intuitively? What shuffled around when composition happened? What does the result of product mean with respect to $\mathrm{\pi <!--\; \pi \; -->}$ and $\mathrm{\sigma <!--\; \sigma \; -->}$?

Solve the following equation in ${\mathbb{Z}}_{107}$:
$101·x-6=4.$

An approximation by Stirling's formula
Let $0<\mathrm{\alpha <!--\; \alpha \; -->}<1$ be a real number and αn be $\mathrm{\alpha <!--\; \alpha \; -->}n$ integer. I want to prove the following fact
$(\genfrac{}{}{0ex}{}{n}{\mathrm{\alpha <!--\; \alpha \; -->}n})=\frac{1+o(1)}{\sqrt{2\mathrm{\pi <!--\; \pi \; -->}\mathrm{\alpha <!--\; \alpha \; -->}(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})n}}{2}^{nH(\mathrm{\alpha <!--\; \alpha \; -->})},$
where $H(\mathrm{\alpha <!--\; \alpha \; -->})=-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}{\log }_2<!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}){\log }_2<!--\; \; -->(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})$. To do this, I've used the Stirling's formula for the factorial:
$n!=(\frac{n}{e}{)}^n\sqrt{2\mathrm{\pi <!--\; \pi \; -->}n}{e}^{r(n)},$,
where $1/(12n+1)<r(n)<1/12n$.
What I've tried:
$$\begin{gathered} (\genfrac{}{}{0ex}{}{n}{\mathrm{\alpha <!--\; \alpha \; -->}n})=\frac{n!}{(\mathrm{\alpha <!--\; \alpha \; -->}n)!((1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})n)!}=\frac{(\frac{n}{e}{)}^n\sqrt{2\mathrm{\pi <!--\; \pi \; -->}n}{e}^{r(n)}}{(\frac{\mathrm{\alpha <!--\; \alpha \; -->}n}{e}{)}^{\mathrm{\alpha <!--\; \alpha \; -->}n}\sqrt{2\mathrm{\pi <!--\; \pi \; -->}\mathrm{\alpha <!--\; \alpha \; -->}n}{e}^{r(\mathrm{\alpha <!--\; \alpha \; -->}n)}(\frac{(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})n}{e}{)}^{(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})n}\sqrt{2\mathrm{\pi <!--\; \pi \; -->}(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})n}{e}^{r((1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})n)}} \\ \frac{e^{r(n)-<!--\; -\; -->r(\mathrm{\alpha <!--\; \alpha \; -->}n)-<!--\; -\; -->r((1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})n)}}{({\mathrm{\alpha <!--\; \alpha \; -->}}^{\mathrm{\alpha <!--\; \alpha \; -->}}(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}{)}^{(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})}{)}^n\sqrt{2\mathrm{\pi <!--\; \pi \; -->}\mathrm{\alpha <!--\; \alpha \; -->}(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})n}}=\frac{e^{r(n)-<!--\; -\; -->r(\mathrm{\alpha <!--\; \alpha \; -->}n)-<!--\; -\; -->r((1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})n)}}{\sqrt{2\mathrm{\pi <!--\; \pi \; -->}\mathrm{\alpha <!--\; \alpha \; -->}(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})n}}{2}^{nH(\mathrm{\alpha <!--\; \alpha \; -->})}. \end{gathered}$$
Now how can I show that $e^{r(n)-<!--\; -\; -->r(\mathrm{\alpha <!--\; \alpha \; -->}n)-<!--\; -\; -->r((1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})n)}=1+o(1)$?

Discrete math: logical equivalent statement and statement forms
"Two statement forms are called logically equivalent if, and only if, they have identical truth values for each possible substitution of statements for their statement variables... denoted $P\equiv <!--\; \equiv \; -->Q$.
Two statements are called logically equivalent if, and only if , they have logically equivalent forms when identical component statement variables are used to replace identical component statements."
Later in the exercise section she writes: $p="x>5"$. Do you not use $\equiv <!--\; \equiv \; -->$ for statement definitions and if so, how do you symbolise equivalence between two statements, $p\equiv <!--\; \equiv \; -->q$ or $p=q$.
Since p and q by them selves could technically be seen as statement forms, is there a difference between $p\equiv <!--\; \equiv \; -->q$ and $p=q$?

Question on Predicate Logic "If Then" and "x if y"
Here is the question I have: “There exists a restaurant that, if it is midnight, then this restaurant would not be open.” and “Not all restaurant are open if it is midnight.” Assume the set of all restaurant as the universe of discourse.
I assume P(x) is the case that the restaurant will not be open and A is the event that it is midnight.
And I got $\exists x(A\to P(x))$ for both sentences. Is there anything I did wrong? Since the question assumes there will be two individual answers for the statements, and have a sub-question requiring me to prove these two logic notation to be logically equivalent.

Pairwise Non Isomorphic 3-regular Graphs of certain orders
I am currently reviewing some graph theory before an exam and ran into the following problem which has me fairly stumped.
The number of pairwise non-isomorphic 3-regular graphs of order 8 is 6, the number of pairwise non-isomorphic 3-regular graphs of order 10 is 21, and the number of pairwise non-isomorphic 3-regular graphs of order 14 is 540. Using this information, determine the number of pairwise non-isomorphic connected 3-regular graphs of order 14.
I know that pairwise non-isomorphic graphs means that no two graphs in the set of graphs for a said order should be the same, i.e. it should have a bijection in each pair of graph's vertex sets.
However, I'm unsure how to utilise the previous orders to determine the number when the order is 14.
Any hints and help regarding this would be appreciated.

Inverse Discrete Fourier Transform equation problem
How can I obtain de inverse discrete Fourier Transform of the following equation: $X(j\mathrm{\omega <!--\; \omega \; -->})=\underset{k=-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(-<!--\; -\; -->1{)}^k\mathrm{\delta <!--\; \delta \; -->}(\mathrm{\omega <!--\; \omega \; -->}-<!--\; -\; -->\frac{k\mathrm{\pi <!--\; \pi \; -->}}{2})$
I have tried by expanding its Fourier series and then doing the inverse and by tables, but I end up obtaining different results, and I am unsure which is the correct one. A step-by-step solution would be very helpful.
Solution 1 (by Fourier Series)
The signal could be written by its Fourier series in this way $X(j\mathrm{\omega <!--\; \omega \; -->})=\frac{1}{\mathrm{\pi <!--\; \pi \; -->}}\underset{k=-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(1-<!--\; -\; -->e^{-<!--\; -\; -->jk\mathrm{\pi <!--\; \pi \; -->}})e^{jk2\mathrm{\omega <!--\; \omega \; -->}}$
Doing the inverse by definition $x(n)=\frac{1}{2\mathrm{\pi <!--\; \pi \; -->}}{\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}}^{\mathrm{\pi <!--\; \pi \; -->}}X(j\mathrm{\omega <!--\; \omega \; -->})e^{j\mathrm{\omega <!--\; \omega \; -->}n}d\mathrm{\omega <!--\; \omega \; -->}$
Obtaining $x(n)=\frac{1}{\mathrm{\pi <!--\; \pi \; -->}}\underset{k=-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(\mathrm{\delta <!--\; \delta \; -->}(n+2k)-<!--\; -\; -->e^{-<!--\; -\; -->jk\mathrm{\pi <!--\; \pi \; -->}}\mathrm{\delta <!--\; \delta \; -->}(n+2k))$
Solution 2 (by table)
$x(n)=\frac{1}{2\mathrm{\pi <!--\; \pi \; -->}}\underset{k=0}{\overset{N-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}(-<!--\; -\; -->1{)}^k{e}^{jkn\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}}$

How to find the number of ways of picking a subset of $\{a,b,c,\dots <!--\; \dots \; -->,z,0,1,2,\dots <!--\; \dots \; -->,9\}$ of two elements?

Find the generating function of $f(n)=\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n}{k})(-<!--\; -\; -->1{)}^{n-<!--\; -\; -->k}{C}_k$
I want to find the generating function of $f(n)=\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n}{k})(-<!--\; -\; -->1{)}^{n-<!--\; -\; -->k}{C}_k$, where $C_k$ is the k-th Catalan number. So, using the definition of an ordinary generating function:
$F(x)=\underset{n\ge <!--\; \ge \; -->0}{\sum <!--\; \sum \; -->}(\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n}{k})(-<!--\; -\; -->1{)}^{n-<!--\; -\; -->k}{C}_k)x^n$
recalling that: $C(x)=\underset{n\ge <!--\; \ge \; -->0}{\sum <!--\; \sum \; -->}{C}_n{x}^n=\frac{1-<!--\; -\; -->\sqrt{1-<!--\; -\; -->4x}}{2x}$
The first idea was to see F(x) as a product of two formal series and I have already seen a proof in this regard where they use the theorem of residues, yet I am looking for something less refined. Any idea?

Alternative way of writing the stars and bars formula where each bar is associated with at least one star.
I was looking for a different way of writing the formula of the number of different k-tuples of non-negative integers whose sum is equal to n and I thought of this formula followed by this combinatorial proof:
$\underset{i=1}{\overset{k}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n+1}{i})(\genfrac{}{}{0ex}{}{k-<!--\; -\; -->1}{i-<!--\; -\; -->1})$
The first combination is the number of positions that we can choose from to place the bars. There are $n+1$ positions to choose from, since now we can also place the bars before and after the stars. The next combination is the number of different ways of splitting the stars between the positions that were chosen for the bars to be placed.
Lastly, we'll have to add all the different ways of combinations of bar positions and number of bars in each position. I haven't found a flaw in my proof yet but I can't seem to conclude that the above formula is equal to
$(\genfrac{}{}{0ex}{}{n+k-<!--\; -\; -->1}{k-<!--\; -\; -->1})$
Could someone more experienced verify my proof or show me where the flaw is?

Let $f:R\to <!--\; \to \; -->R$ be a map of class $C^1$. Show the set $P=\{x\in R:f\prime (x)\ne 0\}$ is open.

Prove the generating function $\underset{N\ge <!--\; \ge \; -->0}{\sum <!--\; \sum \; -->}(\genfrac{}{}{0ex}{}{M+N}{N})x^N=\frac{1}{(1-<!--\; -\; -->x{)}^{M+1}}$