Advanced Math Help with Any Problems!

At least how many residents of the country have the same birthday?
Let's suppose that a country has 11.000.000 people. At least how many residents of the country have the same birthday?
Take into consideration that leap years exist.
What I did is:
Let's suppose we have 2021. I take $\frac{\mathrm{10.000.000}\cdot <!--\; \cdot \; -->2021}{\mathrm{11.000.000}}=1838$.

Number of functions of k-arity
I want to find the number of possible functions $g_i$ acting on k elements $(p_1,\dots <!--\; \dots \; -->,p_k)$ if the possible values of $p_j$s and $g_i$ are $(v_1,\dots <!--\; \dots \; -->,v_n)$.
My thoughts: $g_i:\{v_1,\dots <!--\; \dots \; -->,v_n{\}}^k\to <!--\; \to \; -->\{v_1,\dots <!--\; \dots \; -->,v_n\}$, $g_i:(p_1,\dots <!--\; \dots \; -->,p_k)\mapsto <!--\; \mapsto \; -->g_i(p_1,\dots <!--\; \dots \; -->,p_k)$. It's possible to imagine a tabular with each possible value of the input and so having in total $n^k$ rows. Now the values that $g_i$ can take are again between n option for each row, and so I get $n^{kn}$ possible function. Is that true?

Can't solve $(A\setminus <!--\; \setminus \; -->B)\setminus <!--\; \setminus \; -->C=(A\setminus <!--\; \setminus \; -->C)\setminus <!--\; \setminus \; -->(B\setminus <!--\; \setminus \; -->C)=A\setminus <!--\; \setminus \; -->(B\cup <!--\; \cup \; -->C)$ any help?
I have discrete math exam and i can't quite figure out one example.
$(A\setminus <!--\; \setminus \; -->B)\setminus <!--\; \setminus \; -->C=(A\setminus <!--\; \setminus \; -->C)\setminus <!--\; \setminus \; -->(B\setminus <!--\; \setminus \; -->C)=A\setminus <!--\; \setminus \; -->(B\cup <!--\; \cup \; -->C)$
i tried solving it like this and i got stuck: $(A\setminus <!--\; \setminus \; -->B)\setminus <!--\; \setminus \; -->C=(A\cap <!--\; \cap \; -->\stackrel{¯<!--\; ¯\; -->}{B})\cap <!--\; \cap \; -->\stackrel{¯<!--\; ¯\; -->}{C}$, what next ?

Prove that $n\in <!--\; \in \; -->\mathbb{N}$ and $n\le <!--\; \le \; -->3$, then $2^n<3n+2$.
Prove that $n\in <!--\; \in \; -->\mathbb{N}$ and $n\le <!--\; \le \; -->3$, then $2^n<3n+2$
Need to prove this sentence is true. In this problem, I believe I have to use the exhaustion demo technique. Because there is a specific number of cases, in this case between natural 0 and 3.I've tried to solve it, I came to a conclusion, and apparently it's correct. But I would like someone to look at the strategy used for the resolution and if I am correct about my conclusion.
My resolution method:
Create a table with possible values for n and compare the sentence
$\begin{array}{ccl}n& 2^n<3n+2& \text{result}\\ 0& 1<2& \text{true}\\ 1& 2<5& \text{true}\\ 2& 4<8& \text{true}\\ 3& 8<11& \text{true}\end{array}$
Thus, by exhaustion, it is possible to prove that the theorem is true. For any and all $n\ge <!--\; \ge \; -->0$ and $\le <!--\; \le \; -->3$, it will always be less than $2^n<3n+2$.

Let P(n) be an open statement about $n\in <!--\; \in \; -->\mathbb{N}$. In the weak induction, beside the base case, it is stated:
$\mathrm{\forall <!--\; \forall \; -->}k\in <!--\; \in \; -->\mathbb{N}(P(k)⟹<!--\; ⟹\; -->P(k+1))$ (i)
But, in this page right in the beginning, it states differently, according to my understanding (with $a=1$)
$((\mathrm{\exists <!--\; \exists \; -->}k\in <!--\; \in \; -->\mathbb{N})P(k))⟹<!--\; ⟹\; -->P(k+1)$ (ii)
Are the statements (i) and (ii) equivalent? If yes, why?

How many ways we can choose an increasing and a strictly increasing subsequence of length k from the first n natural numbers?
Let $S:=\{1\dots <!--\; \dots \; -->n\}.$. We want to choose
i) increasing subsequences of S length $k\le <!--\; \le \; -->n.$. and
ii) strictly increasing subsequences of S length $k\le <!--\; \le \; -->n.$.
How many ways can we choose them in i) and ii)? I'm a bit confused where even to start, but detailed answers will be appreciated!

How can I demonstrate a succession with a floor function by induction?
Show that for $n\ge <!--\; \ge \; -->1$ it is satisfied that.
$\lfloor <!--\; \lfloor \; -->\frac{1}{2}\rfloor <!--\; \rfloor \; -->+\lfloor <!--\; \lfloor \; -->\frac{2}{2}\rfloor <!--\; \rfloor \; -->+\lfloor <!--\; \lfloor \; -->\frac{3}{2}\rfloor <!--\; \rfloor \; -->+\cdots <!--\; \cdots \; -->+\lfloor <!--\; \lfloor \; -->\frac{n}{2}\rfloor <!--\; \rfloor \; -->=\lfloor <!--\; \lfloor \; -->\frac{n}{2}\rfloor <!--\; \rfloor \; -->\lfloor <!--\; \lfloor \; -->\frac{n+1}{2}\rfloor <!--\; \rfloor \; -->$
where ⌊x⌋ is the lower integer part of x (floor function evaluated at x).
any idea to start to make it.

Is it true that for every regular language $L\subseteq <!--\; \subseteq \; -->\{0,1{\}}^{\ast <!--\; \ast \; -->}$ the language $\{w^{|w|}|w\in <!--\; \in \; -->L\}$ is also regular?

How many relations are there on a set with n elements?
Discrete Mathematics and Its Applications 8th Edition by Kenneth Rosen states the following on Pg 600:
...a binary relation from A to B is a set R of ordered pairs, where the first element of each ordered pair comes from A and the second element comes from B.
So if A has 3 elements and B has 3 elements, the binary relation should have $3\cdot <!--\; \cdot \; -->3=9$ elements.
Now let us consider the case where we have a binary relation from A to itself. In this case also I expect the binary relation to have $3\cdot <!--\; \cdot \; -->3=9$ elements.
However, on Pg 602 the author states that:
there are $2^{3^2}=2^9=512$ relations on the set (a, b, c).
I am quite confused by this and would appreciate any clarification.

Can substitution be used to solve a single equation that represent a partial order with two variables?
I am still trying to wrap my head around partial orders and my algebra is not the best... My question is pertains to the following example:
ex.) Let a relation R be defined on the set of real numbers as follows: $xRy\iff 2x+y=3$. Prove that this relation is antisymmetric.
Solution: aRb is $(2a+b=3)$, since bRa is $2b+a=3$.
$(2a+b)-(2b+a)=3-3$
$2a+b-2b-a=0$
$a-b=0$
$a=b$
My question is, why subtract the two equations? Do we subtract the two equations because a partial order is a way to quantify how objects in a set are different from each other?And is it possible to solve the same equation by substitution?
I solved: $2x+y=3$
$x=3-<!--\; -\; -->y/2$
and plugged x back into the equation and solved for y.
$2(3-<!--\; -\; -->y/2)+y=3$
$3-<!--\; -\; -->y+y=3$
$3+y=3+y$
$y=y$
I tested a few values to find that the equation $2x+y=3$ is true for any value y that is added to $2(3-<!--\; -\; -->y/2)$

Prove that $f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdot \cdot \cdot +a_{1}x+a_0,x^n=O(f(x))$

Is this function surjective or injective?
I have the following function $f:\mathbb{Z}\times <!--\; \times \; -->\mathbb{N}\to <!--\; \to \; -->\mathbb{Q},f(m,n)=m/n$. Is this function injective or surjective? How can I prove it? I'm new to discrete maths so please bear with me.

Count the number of integer solutions to $x_1+x_2+x_3+x_4+x_5+x_6=25$ when, $x_1,x_2,x_3$ are odd and $x_4,x_5,x_6$ are even and $x_i$ is N.