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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 ${26}^2\times {10}^4$ 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)?

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?

A catering service offers 12 appetizers, 9 main courses, and 6 desserts. A customer is to select 5 appetizers, 3 main courses, and 4 desserts for a banquet. In how many ways can this be done?

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.

Let n be a positive integer. Find the number of permutations of $(1,2,...,n)$ such that no number remains in its original place.

Solution: To do this, first we are going to count the number of permutations where at least one number remains in its place, according to the inclusion-exclusion principle, we must first add the permutations where a given number is fixed, then subtract the permutations where 2 given numbers are fixed and so on.

To find a permutation that fixes $k$ given elements, we only have to arrange the rest, which can be done in $(n-k)!$ ways. However, if we do this for every choice of $k$ elements, we are counting $(\genfrac{}{}{0ex}{}{n}{k})(n-k)!=\frac{n!}{k!}$ permutations. Since in total there are $n!$ permutations we get as our result:
$n!-(\frac{n!}{1!}+\frac{n!}{2!}-\frac{n!}{3!}+...+(-1{)}^{n+1}\frac{n!}{n!})$
I'm a bit confused about a point of the solution, when we fix $k$ elements and rearrange the other $n-k$, some of the remaining elements will stay fixed in their position right? so why does this work?

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:
$\frac{24\cdot 1}{24\cdot 23}=\frac{1}{23}$
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?

Let $f$ be a function from $A\to B$.
$|A|=4,|B|=3$
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?