Explore Discrete Math Examples

Find the maximum of the following function of summation
I ran into this problem:
$$f(\mathrm{\sigma <!--\; \sigma \; -->})=\underset{i=\mathrm{\sigma <!--\; \sigma \; -->}-<!--\; -\; -->1}{\overset{n-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}(\frac{\mathrm{\sigma <!--\; \sigma \; -->}-<!--\; -\; -->1}{i}{)}^2$$
Find the maximum of this function with $1<\mathrm{\sigma <!--\; \sigma \; -->}<n$, $n>4$. My conjecture is that it achieves max with $\mathrm{\sigma <!--\; \sigma \; -->}=\frac{n}{2}+1$ when n is even and $\mathrm{\sigma <!--\; \sigma \; -->}=\frac{n-<!--\; -\; -->1}{2}+1$ when n is odd (according to a simulation). However I cannot prove this. Any hint on what I can do?

Discrete Math on Induction proofs.
Show by induction that the following formulas hold for
$$\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{i}^3=\frac{n^2(n+1{)}^2}{4}$$
Not sure how to go about this problem. Can someone help please?

Discrete Math - Combinatorics - Trinomial Coefficients question
Let $k,l,m,n\in <!--\; \in \; -->Z\ge <!--\; \ge \; -->0$ be such that $n=k+l+m$. The trinomial coefficient $(\genfrac{}{}{0ex}{}{n}{k,l,m})$ is given by the rules:
1. for $k+l=n$, $(\genfrac{}{}{0ex}{}{n}{k,l,0})=(\genfrac{}{}{0ex}{}{n}{k,0,l})=(\genfrac{}{}{0ex}{}{n}{0,k,l})=(\genfrac{}{}{0ex}{}{n}{k})$
2. $(\genfrac{}{}{0ex}{}{n}{k,l,m})=(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->1}{k-<!--\; -\; -->1,l,m})+(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->1}{k,l-<!--\; -\; -->1,m})+(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->1}{k,l,m-<!--\; -\; -->1})$
The following questions use this definition.
(a) What are all the trinomial coefficients for $n=1,2,3$?
(b) Describe the "triangle" of trinomial coefficients (Hint: Think three dimensional Pascal's triangle).
I don't understand the notation at all: n choose k,l,m. What does that even mean? And I don't get how to set up part (a) at all either. It's just very confusing. For (b) I can visualize a Pascal's "pyramid." One where the number below is a sum of the 3 above it. Something like that. But other than that, I'm not really sure what's going on.

Discrete Math: Multiplying a set by $\mathrm{\varnothing <!--\; \varnothing \; -->}$. How would you multiply any set by $\mathrm{\varnothing <!--\; \varnothing \; -->}$? Lets say $A\times <!--\; \times \; -->\mathrm{\varnothing <!--\; \varnothing \; -->}$. Would that simply be equal to $\mathrm{\varnothing <!--\; \varnothing \; -->}$? or Would I write out $(a,\mathrm{\varnothing <!--\; \varnothing \; -->}),(a_1,\mathrm{\varnothing <!--\; \varnothing \; -->}),(a_2,\mathrm{\varnothing <!--\; \varnothing \; -->})$ etc. If possible could you explain your answer in words or short proof?

Discrete Math Clause Count Question
In a CNF formula, a clause contains one or more terms. Each term is either a variable, or the negation of a variable. The only limitations we will place on the clauses (for this problem) is that if a clause mentions a variable (or its negation) once, then it must not mention the same variable (or its negation) again. Therefore, both of the following clauses are disallowed: $(a\vee b\vee a)(a\vee b\vee \neg a)$ However, your program might well generate similar clauses with different negation patterns. The following are four different clauses: $(a\vee b),(a\vee \neg b),(\neg a\vee b),(\neg a\vee \neg b)$
Part (a) - In general, as your program runs, it will generate larger and larger clauses, with a maximum of v terms per clause, if there were v variables mentioned in the original formula. Calculate the number of possible clauses which have exactly n terms, if the original formula mentioned v variables
Part (b) - Calculate the sum total number of clauses (of various sizes) that your program might generate, if the original formula mentioned v variables.
Part (c) - Suppose the original formula mentioned v variables, but every single clause had two or fewer terms. Calculate the total number of clauses which your program might generate.
Part (d) Based on your result from Part (b), give a very rough (thousands, millions, etc) estimate of how many clauses your program might generate if $v=10$, and the clauses in the original formula can be of any size. What would happen if v were to double? How does the count change?

Combinatorics problem involving teams (Discrete math)
A company has 10 men and 18 women. A work team consists of two workers. What is the maximum number of work teams (man-man,woman-woman,man-woman) that can be formed from this group? How many different ways can this maximum number of work teams be formed? I know the max amount of teams is 14. I tried 28 choose 2 for the maximum number of team combinations but it's wrong.

Relation problem.
Let $A=\{1,2,3,4\}$ and let $f:\mathcal{P}(A)\to <!--\; \to \; -->\mathbb{N}$ be the function defined by saying that f(X) is the sum of elements of X, for each $X\in <!--\; \in \; -->\mathcal{P}(A)$. (If $X=\mathrm{\varnothing <!--\; \varnothing \; -->}$, then by convention we say that $f(X)=0$). Define the relation $\sim <!--\; \sim \; -->$ on P(A) as follows:
$X\sim <!--\; \sim \; -->Y$ if and only if $f(X)>f(Y)$ or $X=Y$.
Write down whether ∼ is reflexive, symmetric, antisymmetric, transitive.
The part where I am struggling is to prove reflexivity.
Reflexive: $a\sim <!--\; \sim \; -->a$ for all a. So that means $f(X)>f(Y)$ is false, since $f(X)>f(X)$ is false.
But $X=Y$ part is true since, $X=X$ for all $X\in <!--\; \in \; -->\mathcal{P}(A)$.

Solving an equation involving rearrangements of a 3 digit number
Bob has S amount of money where $100\le <!--\; \le \; -->S\le <!--\; \le \; -->999$ i.e it is a three-digit number. Bob spends some amount for buying lunch which is a rearrangement of the digits of S. Also, the leftover money he has is also a rearrangement of the digits of S. What is the value of S?

Discrete math three fundamental principles
Suppose that there are 15 identical copies of The Great Gastby and 12 distinct biographies on a bookshelf.
(a) How many different selections of 12 books are possible?
(b) How many different selections of 10 books are possible?
I figured out (a)
$2^{12}=4096$
But I'm stuck on b.<br<I tried $2^{10}=1024$ but thats not the answer.

Discrete math - confusion in onto functions
At the CH Company, Joan, has a secretary Teresa, and three other administrative assistants. If seven accounts must be processed, in how many ways can Joan assign the accounts so that each assistant works on at least one account and Teresa's work includes the most expensive account?

Prove that we can get any number by just adding and subtracting distinct exponents of 3

Concrete Mathematics: Formulating definition for value of a general infinite sum
I am having trouble following the explanation which I will reproduce below shortly. I think it is just saying if an infinite sum has a "bounding constant" such as the first example they give, $S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$ bounded by 2 then them resulting sum is finite even if the "inputs" are infinite. If the sum has no bounding constant then the sum result is infinite.
Below is the paragraph to give context so I can highlight the part that confuses me:
Let's try to formulate a good definition for the value of a general sum $\underset{k\in <!--\; \in \; -->K}{\sum <!--\; \sum \; -->}$, where K might be infinite. For starters, let's assume that all the terms ak are nonnegative. Then a suitable definition is not hard to find: If there's a bounding constant A such that
$$\underset{k\in <!--\; \in \; -->F}{\sum <!--\; \sum \; -->}{a}_k\le <!--\; \le \; -->A$$
I think I get the above but then this last part I am not following:
for all finite subsets $F\subset <!--\; \subset \; -->K$, then we define $\underset{k\in <!--\; \in \; -->K}{\sum <!--\; \sum \; -->}{a}_k$ to be the least such A.
Is this a way of saying our sum will be A big if that is the bounding constant? I don't understand what they mean by "the least such A".

if
$$a=\{3n\mid <!--\; \mid \; -->n\in <!--\; \in \; -->{\mathbb{Z}}^{+}\}$$
and
$$b=\{3^{2m}\mid <!--\; \mid \; -->m\in <!--\; \in \; -->{\mathbb{Z}}^{+}\},$$
prove that b is a subset of a.
I think the question is wrong. I think a should be a subset of b.

Discrete Math Question on Set Theory.
Prove, using double inclusion, that if $A=C$ and $B=C$, then $A=B$, whatever the sets A,B and C.
First of all, can someone define double inclusion? And then can he/she prove it?

Discrete Math cardinality finding out
Finding cardinality of the set
For all x less than 100, x/10 is set of Natural number Union 0.
Is the answer 9?

Assume that $a\equiv <!--\; \equiv \; -->b\mathrm{mod}n$ and that n∣a. Show that n∣b
Here is what I have.
If $a\equiv <!--\; \equiv \; -->b\mathrm{mod}n$ this implies $a=b+xn$ for some $x\in <!--\; \in \; -->\mathbb{Z}$. (1)
Since n∣a we can say $a=yn$ for some $y\in <!--\; \in \; -->\mathbb{Z}$. (2)
Combining (1) and (2) we have $yn=b+xn$ (3)
Dividing (3) by n we have that n∣b (4).
Is this correct? Can I divide like this in line (4) in this kind of proof?

Discrete Math: Multiplication Principle Word Problem
Assuming that there will never be more than 500 million people in the United States before the year 2050 AD, how many alphanumeric-upper digits are necessary to provide every person with a personal zip code? (Notice that businesses and institutions are being neglected in this count.)
If the answer is 9 digits, how can I write this in the form $(\genfrac{}{}{0ex}{}{n}{x})$?

Discrete Math on Functions as bijection
Can Someone help me on how to Prove that $f:\mathbb{Z}\times <!--\; \times \; -->\mathbb{Z}\to <!--\; \to \; -->\mathbb{Z}\times <!--\; \times \; -->\mathbb{Z}$ defined by $f(a,b)=(-b,a)$ is a well-defined bijection.

Discrete Math - Proofs and Predicates
Give an example of a domain U and predicates P and Q such that
$(\forall x\in U,P(x))\to (\forall x\in U,Q(x))$ is true but,
$\forall x\in U,(P(x)\to Q(x))$ is false
Any help please on any predicates or domains that work for such logic?

Prove that there exists infinitely many prime numbers of the form $4k+3$ for some integer k.
We shall prove this theorem by contradiction. Assume $p_n$ is the largest prime number of the form 4 $k_n$ +3 for some n $\in <!--\; \in \; -->$ N. Consider an odd value of n. Even if n is even, we could miss one of the prime numbers to get an odd n. Now, consider
$$\begin{array}{rl}N& =2(4k_1+3)(4k_2+3)(4k_3+3)...(4k_n+3)+1\\ & =4m+2\cdot <!--\; \cdot \; -->3^n+1\\ & =4m+2\cdot <!--\; \cdot \; -->(4-<!--\; -\; -->1{)}^n+1\\ & =4m+2\cdot <!--\; \cdot \; -->4a-<!--\; -\; -->2+1\\ & =4m+2\cdot <!--\; \cdot \; -->4a-<!--\; -\; -->1\\ & =4m+3\end{array}$$
for some m,a $\in <!--\; \in \; -->$ N
Two cases are possible. N is definitely not divisible by 4$k_i$+3 i $\in <!--\; \in \; -->$ [1,n].
- All numbers between 4 $k_n$ +3 and N are composite $⟹<!--\; ⟹\; -->$ they are formed from a unique prime factorisation of the finite set of prime numbers. Then N is prime. so contradiction.
- There is a new prime between 4 $k_n$ +3 and N of the form $4k+3$ so N is a multiple of that prime(Note that N cannot be a multiple of $4k+1$ since we have 2 occurring only once in our expression for $N-<!--\; -\; -->1$), which any way is again a contradiction.
My prof said that this proof misses some cases but I am unable to figure out which case. Please help. I feel pretty confident of my proof.