Combinatorics: Examples, Equations, and Practice Problems

What are $Prn$ and $Crn$ in math?

How do you evaluate ${}_{}{}_{}{}^{9}C_3{}_{}{}^{}$?

A certain disease has an incidence rate of 2%. If the false negative rate is 10% and the false positive rate is 1%, compute the probability that a person who tests positive actually has the disease.
For this question, why do we need to use Bayes' Theorem?

Are irrational numbers prime?

The likelihood of rolling two dice and getting the number seven is what?
$$\begin{gathered} A)\frac{7}{12} \\ B)\frac{1}{11} \\ C)\frac{1}{6} \\ D)\frac{2}{3} \end{gathered}$$

A three-person committee is needed to study ways of improving public transportation. Ho many committees could be formed from the eight people on the board of supervisors?

You know that $P(A|B)=P(B|A)$ and at the same time $P(A)+P(B)>1$.Which of the following statements about the events A and B is necessarilytrue?
a)A and B are exhaustive.
b)A and B are independent.
c) A and B form a partition.
d) A and B are mutually exclusive.
e) None of the above.

The supreme court has given a 6 to 3 decisions upholding a lower court; the number of ways it can give a majority decision reversing the lower court is:
Actually in this question, how to get the statement what does the statement "6 to 3 decisions upholding a lower court" mean and then what we have to find ?

Show without using the preceding results * that the probability $p_m(r,n)=n^{-<!--\; -\; -->1}{E}_m(r,n)$ of finding exactly 𝑚 cells empty satisfies
$p_m(r+1,n)=p_m(r,n)\frac{n-<!--\; -\; -->m}{n}+p_{m+1}(r,n)\frac{m+1}{n}(1)$
* The results which I can't use must be
$E_m(r,n)=(\genfrac{}{}{0ex}{}{n}{m})A(r,n-<!--\; -\; -->m)=(\genfrac{}{}{0ex}{}{n}{m})\underset{\mathrm{\nu <!--\; \nu \; -->}=0}{\overset{n-<!--\; -\; -->m}{\sum <!--\; \sum \; -->}}(-<!--\; -\; -->1{)}^{\mathrm{\nu <!--\; \nu \; -->}}(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->m}{\mathrm{\nu <!--\; \nu \; -->}})(n-<!--\; -\; -->m-<!--\; -\; -->\mathrm{\nu <!--\; \nu \; -->}{)}^r$
and by association
$A(r,n+1)=\underset{k=1}{\overset{r}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{r}{k})A(r-<!--\; -\; -->k,n)$
By the way, $A(r,n)$ is equal to $n!S(r,n)$ where $S(r,n)$ are the Stirling numbers of the second kind.

Let $k,p$ be positive integers. Is there a closed form for the sums
$\underset{i=0}{\overset{p}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{k}{i})(\genfrac{}{}{0ex}{}{k+p-<!--\; -\; -->i}{p-<!--\; -\; -->i})\text{, or}$
$\underset{i=0}{\overset{p}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{k-<!--\; -\; -->1}{i})(\genfrac{}{}{0ex}{}{k+p-<!--\; -\; -->i}{p-<!--\; -\; -->i})\text{?}$

Let $S_i$ be the number of permutations $\{a_1,\dots <!--\; \dots \; -->,a_n\}=\{1,...,n\}$, satisfying the following conditions:
1. $a_i=n$ for some $i\in <!--\; \in \; -->\{1,\dots <!--\; \dots \; -->,n\}$
2. $a_1<a_2<\dots <!--\; \dots \; --><a_i$.
Compute $\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{S}_i$.

Bounding $(\genfrac{}{}{0ex}{}{(2d-<!--\; -\; -->1)n-<!--\; -\; -->1}{n-<!--\; -\; -->1})$
Claim: $(\genfrac{}{}{0ex}{}{3n-<!--\; -\; -->1}{n-<!--\; -\; -->1})\le <!--\; \le \; -->{6.25}^n$
1. Why?
2. Can the proof be extended to obtain a bound on $(\genfrac{}{}{0ex}{}{(2d-<!--\; -\; -->1)n-<!--\; -\; -->1}{n-<!--\; -\; -->1})$, with the bound being $f(d{)}^n$ for some function $f$?

Marriage problem says $\{b_1,b_2,\dots <!--\; \dots \; -->,b_n\}$ and $\{g_1,g_2,\dots <!--\; \dots \; -->,g_n\}$ are $n$ boys and $n$ girls respectively. Let $G_i$=set of girls $b_i$ likes. Then a necessary and sufficient condition for a monogamous marriage to happen such that each $b_i$ gets married to a girl in $G_i$ (a compatible matching) is that for any $1\le <!--\; \le \; -->k\le <!--\; \le \; -->n$, and any $1\le <!--\; \le \; -->i_1<i_2<\cdots <!--\; \cdots \; --><i_k\le <!--\; \le \; -->n$, cardinality of $G_{i_1}\cup <!--\; \cup \; -->\cdots <!--\; \cdots \; -->\cup <!--\; \cup \; -->G_{i_k}$ should be bigger than $k$ that is, take any $k$ boys then the total number of girls that these $k$ boys together like should be at least $k$.
There is apparently a version of marriage problem in which one considers both the sets- the set of girls whom each boy likes and set of the boys whom each girl likes. In that case the condition reduces to considering any k boys out of $\lceil <!--\; \lceil \; -->n/2\rceil <!--\; \rceil \; -->$ instead of $n$ above.

The silver urn contains eight black balls and three brown balls. The golden urn contains eight red balls, three yellow balls, and nine blue balls. If I choose a ball from the silver urn and a ball from the golden urn, what is the probability that I get a black ball and a yellow ball? Use 3 decimals

According to the Washington Lottery’s website for its Cherry Blossom Double instant scratch game, the chance of winning a prize on a given ticket is about 17%. Imagine that a person stops at a conveniencestore on the wayhome from work every Monday, Tuesday, and Wednesday to buy a scratcher ticket and plays the game. What is the probability that the player will not win on Monday or Tuesday, but will win on Wednesday?

There are 5 different boxes and 7 different balls.All the balls are to be distributed in the 5 boxes placed in a row so that any box can recieve any number of balls.

How many possible solutions for 6 wires?
//1st set
ABCDEF
GHIJKL
//2nd set
ABCDEF
HGIJKL
//visual layout of situation
0 1
0 1
0 1
0 1
0 1
0 1

Suppose a stack and I want to find the permutations of numbers 1,2,3,...n.
I can push and pop. e.g. if n=2: push,pop,push,pop 1,2 and push,push,pop,pop 2,1
if n=4 I can only get 14 from the 24 permutations using the stack.. Does anyone know any function F(n) that could produce the number of permutations the stack (only one) can produce?
eg f(1)=1
f(2)=2
f(4)=14

Let's say i have a hand of cards.
The number of cards i have in my hand is x > 3.
How many unordered different triples of cards can i form with the cards in my hand?
Example: i have the following cards in my hand: A B C D i could form
A B C
A B D
A C D
B C D
4 different triples can be formed.

Set with $n$ items, $k$ of which have a certain property $k\le <!--\; \le \; -->n$, $m$ items randomly from that set $m\le <!--\; \le \; -->n$, what is the distribution of the number of chosen items having that property? An item can only be chosen once.
expect an answer in the form $\left(\begin{array}{c}f(k,n,m)\\ g(k,n,m)\end{array}\right)$, but I don't know $f(k,n,m)$ and $g(k,n,m)$.