High school probability Uncovered: Discover the Secrets with Plainmath's Expert Help

I am to create a six character password that consists of 2 lowercase letters and 4 numbers. The letters and numbers can be mixed up in any order and I can also repeat the same number and letter as well. How many possible passwords are there?
What I have pieced together so far:
Well, from the fundamental counting principle, we would definitely need ${26}^2\times <!--\; \times \; -->{10}^4$ but obviously this is not all the possibilities since I can rearrange letters and numbers. Since it is a password the order matters so would I try and do a permuation of some sort like ${}^6{P}_2$ since there are 6 slots to try to rearrange 2 objects (letters)?

A stained glass window consists of nine squares of glass in a 3x3 array. Of the nine squares, k are red, the rest blue. A set of windows is produced such that any possible window can be formed in just one way by rotating and/or turning over one of the windows in the set. Altogether there are more than 100 red squares in the set. Find k.
first, there are 8 Isometries of a square.
Identity, three rotations (90,-90,180) four reflections (vertical, horizontal, two diagonal axis). let G be the permutation group, then |G|=8, and I can find fix(g) for every g.
can someone give me a hint of how to proceed from there.

I'm studying the Cartesian product, which is bound to the idea of a binary relation. Even with Cartesian products of several sets, n-ary Cartesian products, we have to think combinatorically as two sets at a time, recursively. Is there any sort of operation where there is, say, a triary operation happening. Obviously, when we have $3-<!--\; -\; -->2-<!--\; -\; -->1$ we have to think of associative rules, forcing subtraction to be a binary operator "step-through" affair. There is no $3\oplus <!--\; \oplus \; -->2\oplus <!--\; \oplus \; -->1$ operator that does something in one fell swoop to all three numbers is there? Lisp has (+ 1 2 3) and grade-school math has vertical addition with the plus-sign and a line under the lowest number, but these are not "internally" non-binary. The only thing that takes "three at once" is, yes, a product-based operator, again, an n-ary Cartesian product. Correct? Another example would be playing poker and being dealt five cards. The cards were shuffled, which is a combinatoric permutation of the stack of cards, which are then dealt into hands. Is there anything from probability (or anywhere) that doesn't start with a shuffle permutation therefore binary, rather, just looks at the five cards coming together non-permutation-, non-binary-wise?

Ministry of Education are inviting tender for four categories promoting the use of IT in education. Each category consists of 5, 4, 3, and 7 projects, respectively. Each project appears on exactly one category. How many possible projects are there to choose from? Explain your answer.
My Answer: $(5+4)+(4+4)+(3+3)+(7+4)=9+8+6+11=34$ possible projects to choose from. I used the sum rule here.
Is this correct?

I know that there is the addition rule of probability, but I want to understand the intuition behind it. Specifically, why does OR signifies addition in probability theory?

I'm missing something here. Let $X=\{(123),(132),(124),(142),(134),(143),(234),(243)\}$, $A_4$ act on $X$ by conjugation (inner automorphisms) and $x=(123)$, then $4=|\mathcal{O}(x)|=|G|/|G_x|=12/|G_x|$. However, $G_x=\{1\}$
What's wrong here?

I came across this question where I was asked to calculate the probability of drawing 2 identical socks from a drawer containing 24 socks (7 black, 8 blue, and 9 green). When solving this question, I tried tackling it using the counting principle directly as opposed to simply applying the combinatorics formula in hopes to get comfortable with that principle. Though, when I looked up the solution at the end of the book, the answer was completely different, and I am not sure why that is.

What I did was as follows: First, I noticed that when we draw the first sock, we have 24 different options to choose from and then only 1 in order to match it with the first one we chose. Then, I divided it by the total number of ways we can draw 2 socks out of 24 and got the expression:
$\frac{24\cdot <!--\; \cdot \; -->1}{24\cdot <!--\; \cdot \; -->23}=\frac{1}{23}$
The solution at the back of the book took care of each case separately (choosing 2 out of 7 blue socks, OR choosing 2 out of 8 blue socks OR choosing 2 out of 9 green socks), which makes sense, but what is wrong with my approach?

I came across the following question in a textbook (bear in mind that this is the only information given)-
There is a 50 percent chance of rain today. There is a 60 percent chance of rain tomorrow. There is a 30 percent chance that it will not rain either day. What is the chance that it will rain both today and tomorrow?
My instinct was to multiply the probabilities together for today and tomorrow to arrive at an answer of 30 percent. However, the answer given is 40 percent (based on taking the addition of the individual probabilities and subtracting their union). Can someone explain to me why the multiplication rule does not apply here? Does it have to do with independence? Bear in mind, I am trying to relearn probability theory from scratch. Thanks.

There's an 80% probability of a certain outcome, we get some new information that means that outcome is 4 times more likely to occur.
What's the new probability as a percentage and how do you work it out?
As I remember it the question was posed like so:

Suppose there's a student, Tom W, if you were asked to estimate the probability that Tom is a student of computer science. Without any other information you would only have the base rate to go by (percentage of total students enrolled on computer science) suppose this base rate is 80%.
Then you are given a description of Tom W's personality, suppose from this description you estimate that Tom W is 4 times more likely to be enrolled on computer science.
What is the new probability that Tom W is enrolled on computer science.

The answer given in the book is 94.1% but I couldn't work out how to calculate it!
Another example in the book is with a base rate of 3%, 4 times more likely than this is stated as 11%.

Let $f$ be a function from $A\to <!--\; \to \; -->B$.
$|A|=4,|B|=3$
The number of surjective functions by applying the principle of inclusion exclusion is given by: $3^4-<!--\; -\; -->(\genfrac{}{}{0ex}{}{3}{1})2^4+(\genfrac{}{}{0ex}{}{3}{2})1^4$.

The rationale is that we begin with the set of all possible functions, and subtract off the functions with one element in the codomain that is not in the range, and add back the functions with two elements in the codomain not in the range.

However, I don't understand why we are adding back $(\genfrac{}{}{0ex}{}{3}{2})1^4$.

If $(\genfrac{}{}{0ex}{}{3}{1})2^4$ already include functions with two elements from the codomain not in the range, then wouldn't it be done already, as that's all the non-surjective functions?

The same three digit prefix is used for all of the telephone numbers in a small town. How many different telephone numbers are possible by changing only the last four digits?

What is the value of 5P4?