Advanced Math Help with Any Problems!

Discrete Math Identity Proof Binomial Coefficients
The question is to prove this identity:
$\underset{k=0}{\overset{m}{\sum <!--\; \sum \; -->}}(-<!--\; -\; -->1{)}^k\left(\begin{array}{c}n\\ k\end{array}\right)=(-<!--\; -\; -->1{)}^m\left(\begin{array}{c}n-<!--\; -\; -->1\\ m\end{array}\right)$ ! where k, m, $n\in <!--\; \in \; -->Z+$

Prove that $X\subset <!--\; \subset \; -->Y⟹<!--\; ⟹\; -->f^{-<!--\; -\; -->1}(X)\subset <!--\; \subset \; -->f^{-<!--\; -\; -->1}(Y)$ .... Is invertibility assumed?
I was able to go ahead and prove this statement by applying the definition of the pre-image. My questions are:
1. Does this statement assume that function is invertible to begin with. Wouldn't a non-invertible function be a counterexample to this statement.
2. Will this statement hold if we replace $\subset <!--\; \subset \; -->$ with equality sign?

Given $p\in <!--\; \in \; -->[0,1],p{n}^{(d+1)/d}\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$ and $np-<!--\; -\; -->\text{log}n-<!--\; -\; -->d\text{loglog}n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$, for a fixed d, show that
$$n(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->1}{d})p^d(1-<!--\; -\; -->p{)}^{n-<!--\; -\; -->1-<!--\; -\; -->d}\le <!--\; \le \; -->o(1),\text{as }n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}.$$.
I’ve been stuck with this for a day. Taking log of the first expression for p would give logn, but I can’t see where that loglogn comes from.

Proving Mathematical Induction
For which nonnegative integers n is $2n+3\le 2^n$? Prove your answer.
I know that we first have to inspect for the basis case and so we find that $n>3$, since $n=1,2,3$ do not hold but 4 plus does, so $n>3.$. But for the Induction step I don't understand how to start it and the textbook solution is unclear, could anyone explain to me what this means?
Textbook solution: For the inductive step assume that P(k) is true. Then, by the inductive hypothesis, $2(k+1)+3=(2k+3)+2<2^k+2.$. But because $k\ge 1,2^k+2\le 2^k+2^k=2^k+1$. This shows that $P(k+1)$ is true.

Give evidence: Let X and Y be random variables with moment generating functions $M_X$ and $M_Y$, respectively, existing in open intervals about 0. Then $F_X(z)=F_Y(z)$ for all $z\in <!--\; \in \; -->\mathbb{R}$ if and only if $M_X(t)=M_Y(t)$ for all $t\in <!--\; \in \; -->(-<!--\; -\; -->h,h)$ for some $h>0$ .

Number theory Discrete Math
prove $gcd(n,n+2)=1$ or 2. What possible values can $gcd(n,n+3)$ have what about $gcd(n,n+4)$C. What about $gcd(n.n+k)$

No man is weak, unless their name is Bob.
From this expression I see that every man who is weak must be called Bob $(W\Rightarrow <!--\; \Rightarrow \; -->B)$, but the expression does not equate to all Bob's are weak $(B\Rightarrow <!--\; \Rightarrow \; -->W)$, as being named Bob is simply the criteria for allocating everyone who isn't called Bob to the set of "all men who are not weak", meaning that the set of "all weak men" is a subset within the set of "all men named Bob".
I just want to make sure that my reasoning for why it can't be $(B\Rightarrow <!--\; \Rightarrow \; -->W)$ is correct, I know that this might be obvious but I just found this confusing initially.

Trying to solve a simple-looking recurrence relation
I have this recurrence relation that I've been trying to solve,
$$g_m=g_{m-<!--\; -\; -->1}+a_m{g}_{m-<!--\; -\; -->2}$$
with $g_1=g_0=1$ and $m\ge <!--\; \ge \; -->0$. Here, $a_m$ is the m-th term of a known sequence. Any ideas?
If it helps, $a_m=\frac{1}{L^2}(m-<!--\; -\; -->1)(r-<!--\; -\; -->m+2)$ where $L^2\in <!--\; \in \; -->\mathbb{Z},L^2\ne <!--\; \ne \; -->0$, $r\in <!--\; \in \; -->{\mathbb{N}}_0$ and $r\ge <!--\; \ge \; -->|L|$, but other than that, L and r are free. So, if I were to be extremely precise, $g_m$ and $a_m$ would actually be $g_{mrL}$ and $a_{mrL}$.

Show that this propositional equivalence is true:
$\neg (𝑟\leftrightarrow <!--\; \leftrightarrow \; -->𝑠)\equiv <!--\; \equiv \; -->(\neg 𝑟\wedge 𝑠)\vee (𝑟\wedge \neg 𝑠)$
My try out was to compare by truth table, but it is not that the exercice is asking. I need the resolution using arguments as "the morgan" and other simplifications as "$\neg (𝑟\leftrightarrow <!--\; \leftrightarrow \; -->𝑠)\equiv <!--\; \equiv \; -->(\neg 𝑟\wedge 𝑠)\vee (𝑟\wedge \neg 𝑠)$."
When I tryed the simplification on the right side, I could reach something like that:
Changing the sides to be easier...
Matching those two sides, in truth table, it doesn't get the same result. The left side gets the result: F V F F; and the right side results in F V V F. Proving that my resolution is wrong. Can someone help me with this argumentation?

I'm struggling with the intuition behind why Chernoff bounds differ in the upper and lower tails. That is, for the lower tail we have:
$Pr(X\le <!--\; \le \; -->(1-<!--\; -\; -->\mathrm{\delta <!--\; \delta \; -->})\mathrm{\mu <!--\; \mu \; -->})\le <!--\; \le \; -->e^{-<!--\; -\; -->\frac{\mathrm{\mu <!--\; \mu \; -->}{\mathrm{\delta <!--\; \delta \; -->}}^2}{2}}$
Whereas for the upper tail:
$Pr(X\ge <!--\; \ge \; -->(1+\mathrm{\delta <!--\; \delta \; -->})\mathrm{\mu <!--\; \mu \; -->})\le <!--\; \le \; -->e^{-<!--\; -\; -->\frac{\mathrm{\mu <!--\; \mu \; -->}{\mathrm{\delta <!--\; \delta \; -->}}^2}{3}}$
Both bounds are found in roughly the same manner, so at a high level, why would they differ in value when we don't know that the distribution is necessarily asymmetric?

Discrete math, problem on combinations
there are 6 undercase letters in a password, how many passwords are there if you have to use at least one 'a'?? I have calculated the total number of passwords which is ${26}^6$ and I have calculated the amount with just one 'a' which is $6\cdot <!--\; \cdot \; -->{25}^5$, i just need don't know how to do it when it comes with the at least one.

Discrete math
What is the coefficient of $x^{10}$ in the expansion of $(5x^2+3{)}^{14}$
Leave answer as an expression rather than number.
Do I have to expand using the pascal triangle 10 times or is there a formula cos they said leave it as a mathematical expression?

What does the notation $\mathcal{F}\subset <!--\; \subset \; -->2^{[n]}$ mean?
What does the highlighted notation mean?
Theorem 1. Let $\mathcal{F}\subset <!--\; \subset \; -->2^{[n]}$ be such that |A| is odd for every $A\in <!--\; \in \; -->\mathcal{F}$ and $|A\cap <!--\; \cap \; -->B|$ is even for every distinct $A,B\in <!--\; \in \; -->\mathcal{F}$. Then $|\mathcal{F}|\le <!--\; \le \; -->n$.

Sum of multiple of two binomial coefficient
I'm trying to show this equality. I try to expand it but I have no idea to go on.Thanks in advance for your help. The equality is
$$\underset{i=0}{\overset{n-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}(2i+1)(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->2}{n-<!--\; -\; -->1-<!--\; -\; -->i})(\genfrac{}{}{0ex}{}{n}{i+1})=2(n-<!--\; -\; -->1)(\genfrac{}{}{0ex}{}{2n-<!--\; -\; -->3}{n})+(\genfrac{}{}{0ex}{}{2(n-<!--\; -\; -->1)}{n})+2(\genfrac{}{}{0ex}{}{2n-<!--\; -\; -->3}{n-<!--\; -\; -->1})$$
I write the left side of equality as below
$$\begin{array}{rl}\underset{i=0}{\overset{n-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}(2(i+1)-<!--\; -\; -->1)(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->2}{n-<!--\; -\; -->1-<!--\; -\; -->i})(\genfrac{}{}{0ex}{}{n}{i+1})& =2\underset{i=0}{\overset{n-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}(i+1)(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->2}{n-<!--\; -\; -->1-<!--\; -\; -->i})(\genfrac{}{}{0ex}{}{n}{i+1})-<!--\; -\; -->\underset{i=0}{\overset{n-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->2}{n-<!--\; -\; -->1-<!--\; -\; -->i})(\genfrac{}{}{0ex}{}{n}{i+1})=2n\sum <!--\; \sum \; -->(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->2}{n-<!--\; -\; -->1-<!--\; -\; -->i})(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->1}{i})-<!--\; -\; -->(\genfrac{}{}{0ex}{}{2(n-<!--\; -\; -->1)}{n})=2n(\genfrac{}{}{0ex}{}{2n-<!--\; -\; -->3}{n-<!--\; -\; -->1})-<!--\; -\; -->(\genfrac{}{}{0ex}{}{2(n-<!--\; -\; -->1)}{n})\end{array}$$

Discrete Math: Counting Problem with balls
1. A bowl contains 10 red balls and 10 blue balls.A woman selects balls at random without looking at them.
a) How many balls must she select to be sure of having at least three balls of the same color?
b) How many balls must she select to be sure of having at least three blue balls?

Proving combinatoric identity using vote casting example.
I'm still having trouble giving a combinatorial proof of this identity using the vote casting example:
$$\underset{k=0}{\overset{m}{\sum <!--\; \sum \; -->}}\left((\genfrac{}{}{0ex}{}{n}{k})\right)=\left((\genfrac{}{}{0ex}{}{n+1}{m})\right),n\ge <!--\; \ge \; -->0$$
To me, the right-hand side represents casting m votes for $n+1$ candidates, since it's a multiset, that seems like we could cast multiple votes for the same candidate. This is equivalent to sum over all the votes each candidate has received, as on the left-hand side.
Is my interpretation correct? I'm still not pretty sure about why the right-hand side has n+1 candidates, while the left side has n.

W denotes a Wiener process in filtration $\mathcal{F}$ , where X is an ${\mathcal{F}}_0$ -measurable
a, Exponential
b, Cauchy
distributed random variable. Let us define
$$\mathrm{\tau <!--\; \tau \; -->}=inf\{t\ge <!--\; \ge \; -->0:\left|W_t\right|\ge <!--\; \ge \; -->X\}.$$ .
What is the probability, that $\mathrm{\tau <!--\; \tau \; -->}$ is finite, so $\mathbf{P}(\mathrm{\tau <!--\; \tau \; -->}<\mathrm{\infty <!--\; \infty \; -->})=?$ And what is $\mathbb{E}\left(\mathrm{\tau <!--\; \tau \; -->}\right)=?$ and $\mathbb{E}(W_{\mathrm{\tau <!--\; \tau \; -->}}\cdot <!--\; \cdot \; -->{\mathrm{\chi <!--\; \chi \; -->}}_{\{\mathrm{\tau <!--\; \tau \; -->}<\mathrm{\infty <!--\; \infty \; -->}\}})=?$ , where ${\mathrm{\chi <!--\; \chi \; -->}}_{\{\mathrm{\tau <!--\; \tau \; -->}<\mathrm{\infty <!--\; \infty \; -->}\}}$ is the indicator variable of the $\{\mathrm{\tau <!--\; \tau \; -->}<\mathrm{\infty <!--\; \infty \; -->}\}$ event?

Use Induction to prove recurrence
Solve the following recurrence and prove your result is correct using induction:
$a_1=0$
$a_n=3a_{n-<!--\; -\; -->1}+4^{n}forn\ge <!--\; \ge \; -->2$

If the probability of frame being lost is P. Then, calculate the mean no. of transmission for the frame to make it success.
Here the probability of frame being lost is P. So the probability of frame reaching safely would be $(1-<!--\; -\; -->P).$
Now lets consider that the frame will reach safely in k-th transmission. That means that the frame being lost $k-<!--\; -\; -->1$ times and reached in k-th time with probability $(1-<!--\; -\; -->P).$. Now a frame requires k-transmissions exactly when the first $k-<!--\; -\; -->1$ attempts fail .... this happens with probability $P^{\text{k-1}}$ and the k-th transmission succeeds , this happens with probability $1-<!--\; -\; -->P.$
For $k=1,$, the probability $=(1-<!--\; -\; -->P)$
For $k=2,$ the probability $=P(1-<!--\; -\; -->P)$
For $k=3,$, the probability = $P^2$ $(1-<!--\; -\; -->P)$ $...........................\mathrm{\infty <!--\; \infty \; -->}$
So the mean number of transmission will be $=(1-<!--\; -\; -->P)+P(1-<!--\; -\; -->P)+$ $P^2$$(1-<!--\; -\; -->P)$..........Which gives me 1. But here mean of the transmission will have to be calculated not the probability mean.So how to calculate the mean of the transmission? But solution saying,
$\underset{k=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}k{P}_k$
$\underset{k=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}k(1-<!--\; -\; -->P)P^{k-<!--\; -\; -->1}$
=$(1-<!--\; -\; -->P)$ $\underset{k=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}k{P}^{k-<!--\; -\; -->1}$
$=(1-<!--\; -\; -->P).$ $\frac{1}{{(1-<!--\; -\; -->P)}^2}$ $=$ $\frac{1}{(1-<!--\; -\; -->P)}$
I don't understand how they multiply $P^{\text{k-1}}$ $(P-<!--\; -\; -->1)$ by k.

If we have $f(r)=r^n$ , where $r^n=(x^2+y^2+z^2{)}^{n/2}$ , and one wishes to examine the level curves of f(r), namely $f(r)=k$ for some real number k, then one might proceed as follows: there would be two cases - (i) take $n>0$ and (ii) $n<0$ .
Now, for case (i), taking $n=1$ , we'll have the upper part of a sphere in 3-space of radius $\sqrt{k}$ . Taking $n=2$ , we'll obtain a whole sphere of radius $\sqrt{k}$ . But what about taking $n>2$ ? That's where I'm getting lost. What kind of a surface are we going to get?
For case (ii), it becomes even more complicated. For example, for $n=-<!--\; -\; -->2$ , we'll have something like a "reversed sphere".
The question is that we can actually view these surfaces in two ways:
1) For example, for $n=5$ , we can arrive at $x^2+y^2+z^2=k^{2/5}$ . For $n=-<!--\; -\; -->2$ , we can arrive at $x^2+y^2+z^2=\frac{1}{k}$ , and both are just spheres in 3-space.
2) We can expand (for $n=5$ ) the expression $(x^2+y^2+z^2{)}^{5/2}=k$ and get a very long expression with powers of 4 and less.
Which approach is "more correct" - (1) or (2)?