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prove inequality by induction - Discrete math
Prove by induction that $\forall n\ge 3$ : $n^2+1\ge 3n$
So I know I need to find my base case, would it be: $n=3$
Then calculate the RHS and LSH
RHS:$3(3)=9$ LHs: $3^2+1=10$ we see that the LHS is greater than or equal to the RHS.
Now for the inductive step:
Assume that the formula is true for an arbitrary $\forall n\ge 3$. We now have to prove $n^2+1\ge 3n$. This is where I get stuck, how can I prove this?
Edit : showing that the assumption applies to $(n+1{)}^2+1\ge 3(n+1)$

Distributing 21 balls in 3 urns with some condition
In how many ways can we distribute 21 distinct balls in 3 different urns such that the first urn contains an even and the second urn contains an odd number of balls? Since 21 can be written only as the sum of three odd numbers or two even numbers and an odd number, we can say the number of distributions of 21 balls in 3 urns is:
$$A_1+A_2+A_3+B$$
where:
$$\begin{gathered} A_i=\text{The number of distributions where the ith urn contains an odd} \\ \text{and the other urns contain an even number of balls} \end{gathered}$$
and:
$$B=\text{The number of distributions where all urns contain an odd number of balls}$$
Now $A_1=A_2=A_3$ (say x) since any state of any of these three can be mapped to the other ones by changing the order of the urns. Now Looks like $B=x-<!--\; -\; -->1$ and I want to prove this by defining a bijective function from the set of all states of the part A1 excluding one particular element to the set of all the states of the part B but the functions I define don't happen to be injective. (Moving one ball from an urn containing an even number of balls to the other one containing an even number of balls could be a possible solution but this function is not injective.) What function do you suggest for this? If we define a bijective function successfully then we'll have:
$$x+x+x+x-<!--\; -\; -->1=3^{21}\Rightarrow <!--\; \Rightarrow \; -->4x-<!--\; -\; -->1=3^{21}\Rightarrow <!--\; \Rightarrow \; -->x=\frac{3^{21}+1}{4}$$
Which is the answer to the problem.

Discrete Math: Unions, Intersections, Complements
Are these answers correct? The union and intersection only include the elements in the universal set?
$U=\{1,2,3,4,5,6,7,8,9,10,11,12\}$ (where U is only a subset of the Universe)
$A=\{2,4,6,8,10,11,12\}$
$B=\{0,1,2,3,4,5,6,7,8,9\}$
$A\cup <!--\; \cup \; -->B=\{1,2,3,4,5,6,7,8,9,10,11,12\}$
$A\cap <!--\; \cap \; -->B=\{2,4,6,8\}$
$(A\cap <!--\; \cap \; -->B{)}^{\prime }=\{1,3,5,7,9,10,11,12\}$
Here is the definition of union given in my book:
$A\cup <!--\; \cup \; -->B=\{x\in <!--\; \in \; -->S|x\in <!--\; \in \; -->A\text{ or }x\in <!--\; \in \; -->B\}$, where A and B are subsets of the universal set S.

Both floor and ceiling functions
$\lfloor <!--\; \lfloor \; -->x-<!--\; -\; -->\lceil <!--\; \lceil \; -->\frac{x}{2}\rceil <!--\; \rceil \; -->\rfloor <!--\; \rfloor \; -->=3.$
If anyone could explain/give me advice on how to solve this question, that would be great.
Edited: I initially equated ceiling(x/2) to n/2, $n/2<x\le <!--\; \le \; -->n/2+1$, but I do not understand how this could be solved with the presence of another x. Another thing that confuses me is how there is a ceiling function inside of the floor function.

logical equivalence statements in discrete math
Construct another English form sentence, which is logically equivalent to that which was given.
"Susan goes to school or Susan does not talk on the phone or Susan does not go to school."

Counting Non-degenerate rectangles Discrete Math
How many non-degenerate rectangles are there of the form $(a\le <!--\; \le \; -->x\le <!--\; \le \; -->b)$ and $(c\le <!--\; \le \; -->y\le <!--\; \le \; -->d)$ where a,b,c and d are integers between (and including) 0 and 99?
Note: A rectangle is degenerate if it has zero area, e.g. $(0\le <!--\; \le \; -->x\le <!--\; \le \; -->1)$ and $(0\le <!--\; \le \; -->y\le <!--\; \le \; -->0)$.
i. Find a formula that's simple but possibly over-counts some rectangles or counts degenerates.
ii. Remove the over-counts from the original formula.
iii. Remove the degenerates from what remains.
iv. Check your formula for a,b,c,d between 0 to 3 by drawing a picture and counting by hand.
Attempt at Solution
There are 100 possible values for x and y, so there are $100\cdot <!--\; \cdot \; -->100=10,000$ rectangles total. I'm not sure how to remove degenerates. It would probably be something like subtracting a combination of all degenerate pairs, pairs that contain $x\le <!--\; \le \; -->1$ or $y\le <!--\; \le \; -->0$.

I have a small pyramid of cards on the table (two cards face down at the bottom, one card face down at the top).
What is the probability of first selecting a red card in the remaining deck of 49, then flipping over one of the lower-level pyramid cards followed by the upper-level pyramid card and hoping that both of these two cards are pairs?

Random bipartite graph with lower bounded degrees
Let i be a positive integer and $V_1,V_2$ be the vertex sets of a bipartite graph. We want to sample uniformly at random a bigraph in the set of bigraphs with vertex set $V_1\cup <!--\; \cup \; -->V_2$ such that for each $v\in <!--\; \in \; -->V_1$, it has at least i neighbors in $V_2$ and for each $u\in <!--\; \in \; -->V_2$, it has at least one neighbor in $V_1$.
Is there any efficient way to do this?

Let A be a nonempty set of real numbers which is bounded below. Let -A be the set of all numbers -x, where $x\in <!--\; \in \; -->A$ . Prove that
$infA=-<!--\; -\; -->sup(-<!--\; -\; -->A)$

Each of the theorems (3.76a)-(3.76e) is called weakening or strengthening, depending on whether it is used to transform the antecedent into the consequent, thus weakening it, or to transform the consequent into the antecedent, thus strengthening it.
What do "to transform the antecedent into the consequent" and "to transform the consequent into the antecedent" mean? My understanding is weakening and strengthening modifies (=transforms) the antecedent and/or the consequent. Why do the authors put it as the antecedent and consequent are transformed one into the other.
$\begin{array}{rl}\text{(3.76) Weakening/strengthening:}& (a)p\Rightarrow <!--\; \Rightarrow \; -->p\vee <!--\; \vee \; -->q\\ & \text{(b)}p\wedge <!--\; \wedge \; -->q\Rightarrow <!--\; \Rightarrow \; -->p\\ & \text{(c)}p\wedge <!--\; \wedge \; -->q\Rightarrow <!--\; \Rightarrow \; -->p\vee <!--\; \vee \; -->q\\ & \text{(d)}p\vee <!--\; \vee \; -->(q\wedge <!--\; \wedge \; -->r)\Rightarrow <!--\; \Rightarrow \; -->p\vee <!--\; \vee \; -->q\\ & \text{(e)}p\wedge <!--\; \wedge \; -->q\Rightarrow <!--\; \Rightarrow \; -->p\wedge <!--\; \wedge \; -->(q\vee <!--\; \vee \; -->r)\end{array}$

Is it possible to solve equlaity between two exponential functions?
And Sal solved the problem by using a table where he calculates the values for each year, which got me thinking about whether there is an easier way to solve it using a formula? I tried writing on a paper:
$$10000+5000n=500\cdot <!--\; \cdot \; -->2^n$$
But didn't know how to solve it, or if it was the right way to do so. I'm still going through Algebra 1 so this might be advanced for me, but I'd like to know whether it can be solved using an easier way or not.

Among the total of 26 members (15 women and 11 men) in the compound be a group of five representatives appointed.
2a)In how many ways can the group be put together, if it is to contain at least two women and at least one man?
b) In how many ways can, if there are two people, Mrs A and Mr B, not both at the same time can be in the group of representatives?
Am I right that it is Binomial theorem but how many ways can I combine these?

Let $X={\mathbb{R}}^2$ . Given $u(\cdot <!--\; \cdot \; -->,\cdot <!--\; \cdot \; -->)$ such that one of the level curves $y=f(x)$ is discontinuous at $x=t$ , construct (or show the existence of) two points $p,l\in <!--\; \in \; -->X$ such that l is a limit point of both $L(p)$ and $U(p)$ but is contained in exactly one of the two

Calculus: $f(n)=n$ for two distinct values of n implies that for some x, $f^{\prime }(x)=1$