Combinatorics: Examples, Equations, and Practice Problems

Two choices: $A$ and $B$.
$P(A)=0.25$; $P(B)=0.75$
There are no conditional probabilities or anything. Each choice is independent of the last.

How many solutions does the following equation have?
$k+l+m+n=30,$
where $k,l,m,n\in <!--\; \in \; -->\mathbb{Z},0\le <!--\; \le \; -->k,l,m,n\le <!--\; \le \; -->10.$

Let there be given an $N\times <!--\; \times \; -->N$ grid; and let $m$ be a natural number.
Let $I$ be the set of all possible ways to place $m$ copies of the letter $X$ in the grid, and $m$ copies of the letter $O$ in the grid (with the restriction that we can only place one letter per space of the grid; in other words, just imagine playing a game of tic-tac-toe on an $N\times <!--\; \times \; -->N$ board for an even number of moves).
Problem:
How many ways are there to completely break the rotational/reflection symmetry of an $N\times <!--\; \times \; -->N$ grid by placing $m$ copies of $X$ and $m$ copies of $O$?

Given is a sequence $⟨<!--\; ⟨\; -->a_1,a_2,\dots <!--\; \dots \; -->,a_n⟩<!--\; ⟩\; -->$ over the alphabet $\{1,2,\dots <!--\; \dots \; -->,m\}$ chosen uniformly at random among the $m^n$ possibilities. What is the expected size of the set $\{a_1,a_2,\dots <!--\; \dots \; -->,a_n\}$?
If $m=n$ it seems the answer tends to $(1-<!--\; -\; -->1/e)n$ as $n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$, but I don't know why.

Let there be a cube with $n$ sides denoted $1,...,n$ each. The cube is tossed $n+1$ times. For $1\le <!--\; \le \; -->k\le <!--\; \le \; -->n$ what is the probability that exactly $k$ first tosses give different number (i.e, the $(k+1)$-st toss give a number that was already gotten.) I really need to know why I got a slightly different answer from the official one.My attempt: Let us build a uniform sample space. $\mathrm{\Omega <!--\; \Omega \; -->}=\{a_i=(i_1,...,i_k)|1\le <!--\; \le \; -->i_j\le <!--\; \le \; -->n\}$. $|\mathrm{\Omega <!--\; \Omega \; -->}|=(n+1{)}^n$, $\mathrm{\forall <!--\; \forall \; -->}\mathrm{\omega <!--\; \omega \; -->}\in <!--\; \in \; -->\mathrm{\Omega <!--\; \Omega \; -->},P(\mathrm{\omega <!--\; \omega \; -->})=\frac{1}{|\mathrm{\Omega <!--\; \Omega \; -->}|}$. We seek for the event $A=\{(i_1,...,i_k,i_{k+1},...,i_{n+1})|i_t\ne <!--\; \ne \; -->i_s,\mathrm{\forall <!--\; \forall \; -->}1\le <!--\; \le \; -->t\ne <!--\; \ne \; -->s\le <!--\; \le \; -->k,k\in <!--\; \in \; -->\{i_1,...,i_k\}\}$ This is the problematic part: $|A|=(\genfrac{}{}{0ex}{}{n}{k})\cdot <!--\; \cdot \; -->k!\cdot <!--\; \cdot \; -->k\cdot <!--\; \cdot \; -->n^{n-<!--\; -\; -->k-<!--\; -\; -->1}$. (Then I and the answer use the formula for probability of an even it a uniform sample space.)The point is, the answer says: $|A|=(\genfrac{}{}{0ex}{}{n}{k})\cdot <!--\; \cdot \; -->k!\cdot <!--\; \cdot \; -->k\cdot <!--\; \cdot \; -->n^{n-<!--\; -\; -->k}$ I don't understand why; First I pick $k$ numbers, count all their permutations, then pick one of them for the $(k+1)$-th toss, and then I have $n-<!--\; -\; -->k-<!--\; -\; -->1$ tosses left, each of which has $n$ possibilities.

Given $n$ distinct objects, there are $n!$ permutations of the objects and $n!/n$ "circular permutations" of the objects (orientation of the circle matters, but there is no starting point, so $1234$ and $2341$are the same, but $4321$ is different).
Given $т$ objects of $k$ types (where the objects within each type are indistinguishable), $r_i$ of the $i^{th}$ type, there are
$\frac{n!}{r_1!r_2!\cdots <!--\; \cdots \; -->r_k!}$
permutations. How many circular permutations are there of such a set?

A committee of 5 is to be chosen from a group of 8 men and 4 women.
Find the probability that
(a) the committee consists of 3 men, 2 women;
(b) the first person chosen for the committee is a man, given that the committee must include exactly 2 women;
(c) the first person chosen is a man, given that the last chosen is a woman.

Originally I was just going to ask the problem on my practice math contest, which is asking how many ways there are to write a nine-digit number containing each digit 1−9 so that the first digit is twice as big as the second and the last two digits are odd. (e.g. 846527913).
I understood that in this case, the number would be in the form of EE(EEEOO)OO or EO(EEOOO)OO where E is an even, O is an odd, the numbers in the parentheses can be ordered in any way, and the only four cases for the first two are 21,42,63, and 84. Calculating this gave me 240, which is definitely wrong (I made a program to check, and it says 7680).
1. How do I solve this particular problem correctly? Am I on the right track?
2. How are these solved in general? Given that it's a 9-digit number and each digit 1−9 is used once, how does one handle constraints such as these?

The problem asks us to calculate:
$\underset{i=0}{\overset{n}{\sum <!--\; \sum \; -->}}(-<!--\; -\; -->1{)}^i(\genfrac{}{}{0ex}{}{n}{i})(\genfrac{}{}{0ex}{}{n}{n-<!--\; -\; -->i})$
The way I tried solving is:
The given sum is the coefficient of $x^n$ in $(1+x{)}^n(1-<!--\; -\; -->x{)}^n$, which is $(1-<!--\; -\; -->x^2{)}^n$ The coefficient of xn in $(1-<!--\; -\; -->x^2{)}^n$ is
$(-<!--\; -\; -->1{)}^{n/2}(\genfrac{}{}{0ex}{}{n}{n/2}).$
Am I doing it right?

Let us suppose to have a set of $N$ elements and to choose $m$ elements out of the set. I know I can do it in $(\genfrac{}{}{0ex}{}{N}{m})$ ways; but I want to know how many of these combinations contain $k$ predefined elements (say $1,2,...,k$).

Let $m,n\ge <!--\; \ge \; -->0$ be two integers. Prove that
$\underset{k=m}{\overset{n}{\sum <!--\; \sum \; -->}}(-<!--\; -\; -->1{)}^{k-<!--\; -\; -->m}(\genfrac{}{}{0ex}{}{k}{m})(\genfrac{}{}{0ex}{}{n}{k})={\mathrm{\delta <!--\; \delta \; -->}}_{mn}$
where ${\mathrm{\delta <!--\; \delta \; -->}}_{mn}$ defined by ${\mathrm{\delta <!--\; \delta \; -->}}_{mn}=\{\begin{array}{ll}1,& \text{if }m=n;\\ 0,& \text{if }m\ne <!--\; \ne \; -->n\end{array}$

How are we able to calculate specific numbers in the Fibonacci Sequence?
I was reading up on the Fibonacci Sequence, $\text{{1,1,2,3,5,8,13,....}}$ when I've noticed some were able to calculate specific numbers. So far I've only figured out creating an array and counting to the value, which is incredibly simple, but I reckon I can't find any formula for calculating a Fibonacci number based on it's position.
Is there a way to do this?

If 2 people out of 7 refuse to sit next to each other in a row of 7 seats, how many ways are possible?

In how many ways will the sum of the number of 1’s and 2’s (total of occurrences of 1's and 2's) equal the sum of the number of 3’s, 4’s, 5’s and 6’s, after n rolls, assuming n = 2i for some positive integer i? What is the probability of this occurring?

A common way to define a group is as the group of structure-preserving transformations on some structured set. For example, the symmetric group on a set $X$ preserves no structure: or, in other words, it preserves only the structure of being a set. When $X$ is finite, what structure can the alternating group be said to preserve?
As a way of making the question precise, is there a natural definition of a category $C$ equipped with a faithful functor to $\text{FinSet}$ such that the skeleton of the underlying groupoid of $C$ is the groupoid with objects $X_n$ such that $\text{Aut}(X_n)\simeq <!--\; \simeq \; -->A_n$?

Suppose you have 100 coins. 96 of them are heavy and 4 of them are light. Nothing is known regarding the proportion of their weights. You want to find at least one genuine (heavy) coin. You are allowed to use a weight balance twice. How do you find it?
Assumptions:
Heavy coins all have the same weight; same for the light coins.
The weight balance compares the weight of two sides on the balance instead of giving numerical measurement of weights.

How many combinations of tests would be there for example, if
$a$ can take values from $1$ to $m$
$b$ can take values from $1$ to $n$
$c$ can take values from $1$ to $p$
$a$, $b$ and $c$ can take m, n and p distinct values respectively. What are the total number of pairwise combinations possible?
With a pairwise testing tool that I am testing, I am getting $40$ results for $m=n=p=6$. I am trying to mathematically understand how I get $40$ values.

Data set $2334444$
I want to find the number of unique numbers of $3$ digit numbers that can be formed using this.
I was thinking of doing $\frac{{}^7{P}_3}{4!\times <!--\; \times \; -->2!}$, but this doesn't seem right.

The question gives us 3 vowels (A,E,O) and 4 consonants (B,C,D,F).
a) How many ways can you make a 7 letter word if each letter can only be used once. (Word doesn't have to be real). This was easy, as the answer is just 7!
b) If the vowels have to be together and the consonants have to be together?
c) If the vowels have to be together?
d) If B and C have to be together, but no other vowels or consonants can be together?

You have 2 boxes that contains 100 balls each, and the balls can be white, blue, red, or half red and blue.
Box A contains: 90 white balls, 10 red balls
Box B contains: 86 white balls, 4 red balls, 9 blue balls, 1 red and blue ball
If you don't know which box you pick a ball from (the probability is equal for what box you chose), what is the probability for pulling exactly 2 white balls and 1 red ball, if you don't put it back and don't take a look at the balls until all 3 are pulled from the box you chose? (You only pick 3 balls).
I thought this would be pretty straight forward by using combinatorics:
P(1 red and 2 white balls) = $\frac{(\genfrac{}{}{0ex}{}{176}{2})}{(\genfrac{}{}{0ex}{}{200}{3})}$