Explore Discrete Math Examples

Here is an equivalence relation $R=\{(x,y)|x-<!--\; -\; -->y\}$ is an integer}
My question is: what is the equivalence class of 1 for this equivalence relation?
Can say indicate the equivalence class of 1 as $[(1)]=\{(x,y),x-<!--\; -\; -->y=\}$
I am confused about how to write the right hand side? can someone help me?

In how many ways can we distribute 20 balls into 5 distinct boxes when it is required that no box remain empty and box number 2 will have no more than 10 balls?
Can someone explain me the steps to answer this question?
The final answer should be 19choose4 - 9choose4, if you can explain why it would really help.

The board of directors of a pharmaceutical corporation has 10 members. Three members of the board of directors are physicians. How many different slates consisting of a president, vice president, secretary, and treasurer be made with exactly one physician. Note that there are 3 physicians out of the 10 directors.

Statistics with discrete math
An upper-level math class has 13 students: 4 of them are females. Two of the females and four of the males live off campus. How many ways can a project group of three students be chosen so that it has two on-campus members and at least one female? Explain your answer

Discrete math logic question
I have the following two questions.
1. For all real numbers x, there is a real number y such that $2x+y=7$ would this be true or false?I think true because if you put
$2(7)+y=14$
$2(8)+y=14$ there will always be a specific y that will make it work is this logic correct.
1. There is a real numbers x that for all real number y, $2x+y=7$ will be true.
would this be false because if you say $x=6$
then you get
$2(6)+2=14$
only if $y=2$ would it work but it would not work for every y.

Find a explicit formula for the recursion
$\begin{array}{rl}{a}_1& =3\\ a_n& =5a_{n-<!--\; -\; -->1}+12\end{array}$
So i found
$\begin{array}{rl}{a}_1& =3\\ a_2& =27\\ a_3& =147\\ a_4& =747\end{array}$

Divisibility Discrete Math
For all integers a, b, c, if $a|(b+c)$, then a|b and a|c True or false?
Im assuming it's false because if you make $a=2,b=3$ and $c=4$, it won't work

Discrete Math - Combinatorics
A computer password is required to be 9 characters long. How many passwords are possible if the password requires 3 letter(s) followed by 6 digits (numbers 0-9), where no repetition of any letter or digit is allowed?
I thought about using the formula: $p(n,r)=n!/(n-<!--\; -\; -->r)!$
So p(9,9) is wrong.. Not sure what to do for this problem..

Bijective function from ${\mathbb{Z}}^{+}$ to $\mathbb{Z}\mathrm{\setminus <!--\; \setminus \; -->}\{-<!--\; -\; -->1,0,1\}$.
I am supposed to give a bijective function from ${\mathbb{Z}}^{+}$ to $\mathbb{Z}\mathrm{\setminus <!--\; \setminus \; -->}\{-<!--\; -\; -->1,0,1\}$, where ${\mathbb{Z}}^{+}$ is the set of positive integers and $\mathbb{Z}$ is the set of integers. I don't quite understand how a set with a smaller size can be surjective when mapped to a set with a larger size; wouldn't there be too many elements to map to? Can anyone help explain the problem and potentially provide a function that meets the criteria?

Form of the smallest vertex cover in a bipartite graph
I'm trying to write a proof of Konig's theorem using Menger's theorem. However, I got stuck along the way. In order to move forward I'd (apparently) need to show the following fact.
Let G be a bipartite graph with partite sets X and Y. The smallest vertex cover of G is of the form $(X-<!--\; -\; -->A)\cup <!--\; \cup \; -->N(A)$ for some $A\subseteq <!--\; \subseteq \; -->X$.
I tried doing a proof by contradiction but to no avail. Maybe the thesis I'm trying to show is false in the first place? Any help would be greatly appreciated.

Find out whether this sequence can be obtained in this particular order:
Each number from 1 to 2021 occurs twice (1,1,2,2,3,3,....so on). Can we arrange it in such a way that between first occurrence and second occurrence of k there are exactly k numbers?
I can see that numbers 1,1,2,2,3,3 can be arranged like 312132. But I am not able to proceed any further. Please help me here as early as possible. Thank you in advance.

In how many ways can you sit 5 people in a row of 20 seats if no 2 can sit together?
I've seen the simpler problem of just sitting 2 people in non consecutive seats. In that case, I would subtract from the total number of ways to sit the 2 persons the number of ways of sitting them together.
In this harder version of the problem,I've though of the same thing, but now considering the case were 2, 3, 4 or 5 sit together. But that seems to count duplicate cases.

Linear congruences - system of two equations with two variables - solutions don't satisfy
$9x+27y\equiv <!--\; \equiv \; -->3(\mathrm{mod}102)$
$16x+4y\equiv <!--\; \equiv \; -->2(\mathrm{mod}102)$
Here's what I did. I subtracted the first equation from the second to get
$9x+27y\equiv <!--\; \equiv \; -->3(\mathrm{mod}102)$
$7x-<!--\; -\; -->23y\equiv <!--\; \equiv \; -->-<!--\; -\; -->1(\mathrm{mod}102)$
After that, I multiplied the first equation by 7. Next, I added to it the second equation multiplied by -9 to eliminate x from the first equation:
$I=II\cdot <!--\; \cdot \; -->-<!--\; -\; -->9+I$
The first equation now looks like this:
$396y\equiv <!--\; \equiv \; -->30(\mathrm{mod}102)$
I checked the solutions online, and this congruence has six solutions less than m which are: $y=6,23,40,57,74,91$

System of three linear congruences with three variables
I have the following system
$\{\begin{array}{c}5x+20y+11z\equiv <!--\; \equiv \; -->13(\mathrm{mod}34)\\ 16x+9y+13z\equiv <!--\; \equiv \; -->24(\mathrm{mod}34)\\ 14x+15y+15z\equiv <!--\; \equiv \; -->10(\mathrm{mod}34)\end{array}$
I'm still fairly new to this so I would like anyone so verify my solution.
First, I multiplied equations 2 and 3 by 5, it's a regular transformation because 5 and 34 are coprime. After that, the coefficient next to x in the second equation is 80, and the coefficient of x in the third equation is 70, both of which are a multiply of 5 so we can eliminate them using the first equation.
Hence we get $\{\begin{array}{c}5x+20y+11z\equiv <!--\; \equiv \; -->13(\mathrm{mod}34)\\ -<!--\; -\; -->275y-<!--\; -\; -->111z\equiv <!--\; \equiv \; -->-<!--\; -\; -->88(\mathrm{mod}34)\\ -<!--\; -\; -->245y-<!--\; -\; -->161z\equiv <!--\; \equiv \; -->-<!--\; -\; -->198(\mathrm{mod}34)\end{array}$
Next, I multiplied the third equation by 55. It is a regular transformation because 55 and 34 are coprime. The coefficient of y in the third equation is -13475, which is a multiple of -275, so we can eliminate it by multiplying the second equation by -49 and adding it to the third equation.
Now we are left with $-<!--\; -\; -->3416\equiv <!--\; \equiv \; -->-<!--\; -\; -->6578(\mathrm{mod}34)$. If we multiply this equation by -1 and reduce the coefficients (because they are larger than the modulus) we get $16z\equiv <!--\; \equiv \; -->16(\mathrm{mod}34)$
We have two typical solutions, $z=1$, $z=18$.
Now, I proceeded to plug in both values of z gradually and I got two solutions:
$x\equiv <!--\; \equiv \; -->22(\mathrm{mod}34)$, $y\equiv <!--\; \equiv \; -->15(\mathrm{mod}34)$, $z\equiv <!--\; \equiv \; -->1(\mathrm{mod}34)$
and $x\equiv <!--\; \equiv \; -->5(\mathrm{mod}34)$, $y\equiv <!--\; \equiv \; -->32(\mathrm{mod}34)$, $z\equiv <!--\; \equiv \; -->18(\mathrm{mod}34)$
NOTE: I omitted the process of plugging both values of z one by one into the equations because I'm fairly sure I know how to proceed from there. I'm interested in knowing if my method of reducing the system to one equation with one variable is correct or not.
EDIT: After plugging in both sets of solutions, I get that both sets satisfy equations (1) and (2), but in both cases of solutions I get that the third equation ends up being $4\equiv <!--\; \equiv \; -->10(\mathrm{mod}34)$ which is false. Where did I go wrong?

I'm self studying How to Prove book and have been working out the following problem in which I have to analyze it to logical form:
Nobody in the calculus class is smarter than everybody in the discrete math class
Now, this is how, I started solving it:
¬(Somebody in the calculus class is smarter than everybody in the discrete math class) ¬(If x is in calculus class then x is smartert than everybody in the discrete maths class)
$C(x)=x$ is in calculus class. $D(y)=y$ is in discrete class. $S(x,y)=x$ is smarter than y
$\neg \mathrm{\exists <!--\; \exists \; -->}x(C(x)\to <!--\; \to \; -->\mathrm{\forall <!--\; \forall \; -->}y(D(y)\wedge S(x,y)))$
But this is the solution given in the Velleman's book:
$\neg \mathrm{\exists <!--\; \exists \; -->}x[C(x)\wedge \mathrm{\forall <!--\; \forall \; -->}y(D(y)\to <!--\; \to \; -->S(x,y))]$
I cannot understand how that answer is correct. Can someone explain the thing I'm missing there ?

Counter-example to $(R\cup <!--\; \cup \; -->S{)}^{\ast <!--\; \ast \; -->}=R^{\ast <!--\; \ast \; -->}\cup <!--\; \cup \; -->S^{\ast <!--\; \ast \; -->}$.
We must show if the following equation involving two relations on a set A is true or false:
$(R\cup <!--\; \cup \; -->S{)}^{\ast <!--\; \ast \; -->}=R^{\ast <!--\; \ast \; -->}\cup <!--\; \cup \; -->S^{\ast <!--\; \ast \; -->}$
Where R∗ indicates the complement relation of R on the same set.
Is the following counter-example correct?
$A=\{a,b\}$.
$R=\{(a,a),(b,b)\}$ so that $R^{\ast <!--\; \ast \; -->}=\{(a,b),(b,a)\}$
$S=\{(a,b),(b,a)\}$ so that $S^{\ast <!--\; \ast \; -->}=\{(a,a),(b,b)\}$
Then $R^{\ast <!--\; \ast \; -->}\cup <!--\; \cup \; -->S^{\ast <!--\; \ast \; -->}$ is $A\times <!--\; \times \; -->A$ while $(R\cup <!--\; \cup \; -->S{)}^{\ast <!--\; \ast \; -->}$ is the empty set, so the equality does not hold.
Would this be a correct counter-example? Thanks in advance.

Generating function of $f(n)=C_n-<!--\; -\; -->\underset{k=1}{\overset{n-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n}{k})f(n-<!--\; -\; -->k)$.

Existence of “sufficiently rich” unions
Let m be a positive integer and define $M\equiv <!--\; \equiv \; -->\{1,\dots <!--\; \dots \; -->,m\}$. Let $A_1,\dots <!--\; \dots \; -->,A_m$ be (not necessarily disjoint and potentially empty) finite sets. Moreover, let $c_1,\dots <!--\; \dots \; -->,c_m$ be non-negative integers. For each $i\in <!--\; \in \; -->M$, $A_i$ may or may not contain precisely $c_i$ elements, but assume that the following is true (here, # denotes cardinality):
$\mathrm{\#<!--\; \#\; -->}\left(\underset{i\in <!--\; \in \; -->M}{\bigcup <!--\; \bigcup \; -->}{A}_i\right)=\underset{i\in <!--\; \in \; -->M}{\sum <!--\; \sum \; -->}{c}_i.$
Conjecture: There exists a non-empty index subset $T^{\ast <!--\; \ast \; -->}\subseteq <!--\; \subseteq \; -->M$ with the following properties:
- $\mathrm{\#<!--\; \#\; -->}\left(\underset{i\in <!--\; \in \; -->T}{\bigcup <!--\; \bigcup \; -->}{A}_i\right)\ge <!--\; \ge \; -->\underset{i\in <!--\; \in \; -->T}{\sum <!--\; \sum \; -->}{c}_i$ whenever $T\subseteq <!--\; \subseteq \; -->T^{\ast <!--\; \ast \; -->}$; and
- $\mathrm{\#<!--\; \#\; -->}\left(\underset{i\in <!--\; \in \; -->T}{\bigcup <!--\; \bigcup \; -->}{A}_i\right)\ge <!--\; \ge \; -->\underset{i\in <!--\; \in \; -->T}{\sum <!--\; \sum \; -->}{c}_i$ whenever $T\supseteq <!--\; \supseteq \; -->T^{\ast <!--\; \ast \; -->}$.
In words, I am looking to establish the existence of an index set $T^{\ast <!--\; \ast \; -->}$ such that if the index set T is either a subset or a superset of $T^{\ast <!--\; \ast \; -->}$, then the union of the $A_i$'s over T contains at least as many elements as the sum of the $c_i$'s over T.
Any references to a proof or hints about a counterexample would be appreciated.