Explore Discrete Math Examples

How to prove that subset at odd size is equal to subset at even size?
I'm trying to prove that:
$$\underset{i=0}{\overset{\lceil \frac{n}{2}\rceil }{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n}{2i})=\underset{i=0}{\overset{\lceil \frac{n}{2}\rceil }{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n}{2i+1})$$
if n is odd, I can do it with the identity: $(\genfrac{}{}{0ex}{}{n}{k})=(\genfrac{}{}{0ex}{}{n}{n-<!--\; -\; -->k})$, because if we have odd number like 7 we have 4 ($=\lceil \frac{n}{2}\rceil$) pairs: (7,0),(6,1),(5,2),(4,3). The left element is odd and the right one is even. So:
$$(\genfrac{}{}{0ex}{}{7}{0})=(\genfrac{}{}{0ex}{}{7}{7}),(\genfrac{}{}{0ex}{}{7}{6})=(\genfrac{}{}{0ex}{}{7}{1}),...$$
But when I'm trying to do this about even numbers (like 8), I can't use this method.
So my question is: What can I do at cases of even numbers?

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.

Proof by cases - discrete math
I need to prove by the cases technique that:
$Ifa,b\in <!--\; \in \; -->Z,anda=b^3,\text{there is}c\in <!--\; \in \; -->Zsoa=9\cdot <!--\; \cdot \; -->cora=9\cdot <!--\; \cdot \; -->c+1ora=9\cdot <!--\; \cdot \; -->c-<!--\; -\; -->1.$

Discrete Math problem combinations and restrictions
I am trying to get my head around an idea but I can't seem to get it to work.
Imagine you have the word "MAMMAL"
Lets see I wanted to figure out how many ways I could rearrange the letters. Well that is easy. It is simply
$6!/3!2!=60$ possibilities
But what if I added a restriction to it such that all M's must be together.
Then I could consider all M's as one letter and it would be $4!/2!=12$ possibilities.
Again I understand that. But what if the restriction was there needs to be a minimum of 2 M's together at all times.
If I were to do the same process as the previous example, I would combine 2 M's into one. Which then would become $5!/2!=60$.
This seems to be a wrong answer because it is the same as my first calculation of finding all possibilities without any restrictions. Can anyone please explain to me as to how I need to approach the last problem of finding number of combinations where at least 2 M's are always together?

Discrete Math Construct Tree from Weights
Consider the weights: 10, 12, 13, 16, 17, 17.
(a) Construct an optimal coding scheme for the weights given by using a tree.
(b) What is the total weight of the tree you found in part (a)?

Upperbound on harmonic number discrete math
How do I show $H_{2^k}\le <!--\; \le \; -->k+1$ for each $k\ge <!--\; \ge \; -->0$?

Discrete Math
I am trying to understand a question I got:
Let U be a set.
For every $A\subseteq <!--\; \subseteq \; -->U$ we define the function: $f_A:P(U)\to <!--\; \to \; -->P(U)$ such as fA(B)=(A∩B).
Prove: f is injective and surjective if and only if $A=U$
First, I do not understand what $f_A$ and $f_A(B)$ mean.
I think that if I will understand this, I will be able to start solving the question.
Any tips will be wonderful!

Arranging books in bookshelves with the capacity of each shelf given
There are k identical bookshelves in which each shelf cannot contain m or more books. In how many ways can n distinct books be arranged on these k bookshelves?
If there is no condition on the capacity of each shelf, the number of ways to arrange books equals $\underset{i\in <!--\; \in \; -->[k]}{\sum <!--\; \sum \; -->}L(n,i)$, where L(n,i) denotes the Lah number. However, because of that constraint, I have trouble in solving the problem.
I tried several ways to solve this problem by separating the cases via (1) the number of shelves which contains the full-number of books, or (2) the number of non-empty shelves. For the second trial, I observed that, if j denotes the number of non-empty shelves, then the number of ways to arrange the books is zero if $j<\lfloor <!--\; \lfloor \; -->n/m\rfloor <!--\; \rfloor \; -->$.
However, these methods does not proceed quite well, since it looks like these methods result in the recurrence relation rather than the exact form of the number.
For the related concepts, I have studied Catalan number, (both signed and unsigned) first and second Stirling number, Bell and Lah number, and the integer partition.
Any insight or comment are welcomed.

Discrete math one to one question
Use the pigeonhole principle to prove that if $n\ge <!--\; \ge \; -->2$ people go to the same party, there are at least 2 people who shake hands with the same number of other people.
Hint: Take the set of people at the party as your domain, define a function that evaluates how many people each person shook hands with.

Help with summations for discrete math
I could start them but I have no idea how to simplify after expanding them.
$$\underset{i=28}{\overset{n}{\sum <!--\; \sum \; -->}}(3i^2-<!--\; -\; -->4i+(5/7^i))$$
$$\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{i}{2^{n-<!--\; -\; -->i+1}}$$

How can i prove this?
$\mathrm{\forall <!--\; \forall \; -->}n\mathrm{\exists <!--\; \exists \; -->}p(p^2\le <!--\; \le \; -->n<(p+1{)}^2)$
The domain of quantifiers is N.

What is a strongly discrete sequence?
What is the definition (and possibly examples or applications) of strongly discrete sequence? I have seen this term, but I am not able to find anything about it.

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.

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.

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.