Combinatorics: Examples, Equations, and Practice Problems

If I create a $10$ digit password with the following requirements:
At least one uppercase letter A-Z: $26$
At least one lowercase letter a-z: $26$
At least one digit from $0-<!--\; -\; -->9$: $10$
At least one common symbol ($(\mathrm{\#<!--\; \#\; -->},\mathrm{\$<!--\; \$\; -->},\mathrm{\%<!--\; \%\; -->},$ etc): $32$
By inclusion-exclusion, I have ~ $3.2333\mathtt{E}+19$ possible combinations
However, if I change one of the requirements to at least two digits $0-<!--\; -\; -->9$, how can I calculate the possible combinations?

Combinatorics select 3 cards out of 52, probability all 3 spades and not.
You get 3 cards out of 52, the order doesn't matter.
a) What is the probability all 3 cards are spades?
b) What is the probability none of the 3 cards are spades?
a) I think it is
$\frac{(\genfrac{}{}{0ex}{}{13}{3})}{(\genfrac{}{}{0ex}{}{52}{3})}\approx <!--\; \backslash approx\; -->0.012941$
b) I know it is
$=\frac{(\genfrac{}{}{0ex}{}{39}{3})}{(\genfrac{}{}{0ex}{}{52}{3})}\approx <!--\; \backslash approx\; -->0.41353$
I know $P(A^c)=1-<!--\; -\; -->P(A)$
And if a is $A$ and b is $A^c$ the equation above is not correct..

I would like to know how to calculate the amount of numbers between 1000 and 100000 that do not contain 2,7 or 8. I had this question in my exam today and, since I have never been able to do probability and combinatorics properly, it's very likely I got it wrong. Could you please show me how to calculate it?
What I thought was: for 4 digit integers, calculate the total amount of possible numbers. Then calculate all numbers which do not contain 2,7 or 8 and deduce that from the total amount of 4 digit numbers. Repeat process for 5 digit integers and sum them up.

A tea shoppe serves 12 different flavors of tea. 4 customers each order a cup of tea. Assuming that each customer's choice is completely random and is independent of any other customer's selection:
1) what is the probability that 3 different flavors are selected by the 4 customers (e.g:: if letters A-L represent flavors, AABC)?
2) what is the probability that only 2 different flavors are selected by these 4 customers (e,g: AABB or AAAB)? 3) what is the probability that 2 students choose 1 flavor and the other 2 choose another flavor (e.g: AABB)?

Finding the power series of a rational function
In many combinatorial enumeration problems it is possible to find a rational generating function (i.e. the quotient of two polynomials) for the sequence in question. The question is - given the generating function, how can we find (algorithmically) the values of the sequence, i.e. the coefficients of the corresponding power series?
I know that for a rational generating function, the sequence satisfies a recurrence relation given by the coefficients of the polynomial in the denominator, so it's really just the question of finding the finite initial values.

If you are a voter in a state with a million other active voters, what is the probability that your vote will be pivotal in the next election?
Assuming a given voter is equally likely to vote for either party, it seems that the answer is $\left(\begin{array}{c}1,000,000\\ 500,000\end{array}\right)\ast <!--\; \ast \; -->(\frac{1}{2}{)}^{500,000}\ast <!--\; \ast \; -->(\frac{1}{2}{)}^{500,000}$ and when I plug it into a calculator it gives me 7.9788E-4, which doesn't seem plausible. Is there something wrong with my reasoning?

There are 400 dormitory rooms along with 400 distinct room numbers for 60 math students. Any math student can stay in any of the 400 dormitory rooms.
Suppose you are a math student and are recording the room number for all math students, one at a time until you have found a match (This means that a room number has already been recorded).
i) What is the probability that it takes more than 30 students for this to happen?
ii) What is the probability that is takes exactly 25 people for this to happen?
Suppose you are a math student and are recording the room number for all math students, one at a time you have found a student who shares the same dormitory room with you. What is the probability that it takes exactly 28 math students for this to happen?
Well, I think that for 1(i), I may first select 30 persons out of 400 students as the numerator, then all those 30 students have 400 choices, so the denominator would be 400^30.
As for 1(ii), I just did it in the same way as (i) but to change 30 into 25. I also multiply 25 since 1 out of 25 has the same choice.
As for 2, I think that everyone just make a different choice than me so i put (399/400)^28 as my answer.

From a deck of 52 cards, we select the ace of spades, plus 4 cards among the 51 remaining cards.
a) What is the expected number of aces among the five selected cards ?
b) We shuffle the five cards and pick one at random. What is the probability that it is an ace ?

Calculate Password Possibilities With Specific Character Pattern
So i need to estimate the number of possibilities of a 8 characters password all stick together with a specified pattern wish have only:
uppercase letter and numbers. no special characters
The pattern is : Number, Number, Up Letter, Up Letter, Number, Up Letter, Number, Up Letter

The problem is as follows: We have box of the following:
- {A} - 40W light bulbs, N(A) = 4
- {B} - 65W light bulbs, N(B) = 5
- {C} - 75W light bulbs, N(C) = 6
We select 2 bulbs at random. Given that at least one of these is rated 75W, what is the probability that they are both rated 75W?
Using a probability tree, I was able figure out the answer by dividing the probability of both bulbs being 75W (6/15)*(5/14) by all the cases that included a 75-W bulb including the case where both were 75 W.
However, how should I do this using combinatorics?

For all positive integer $n\ge <!--\; \ge \; -->8$,
$\lfloor <!--\; \lfloor \; -->\sqrt{n!}\rfloor <!--\; \rfloor \; -->=\lfloor <!--\; \lfloor \; -->\sqrt{n!+2}\rfloor <!--\; \rfloor \; -->,$
where the symbol $\lfloor <!--\; \lfloor \; -->x\rfloor <!--\; \rfloor \; -->$ indicates the floor function.

Let's say I have 3 variables (X, Y and Z) each one can assume a positive integer value, including zero (0, 1, 2, ..., 100), but the sum of the three must be 100. How many combinations do I have ? Different variables can have the same value.

Is there a natural way to represent all the partitions of an integer set $\{1,2,3,...,n\}$ as a matrix in the similar way permutations can be mapped to group of matrices?

Shuffle a standard deck of cards and cut it into three piles. What is the probability that a face card will turn up on top of one of the piles?
There are 12 face cards (four jacks, four queens and four kings) in the deck.

What is $P(EF)$? Is this $P(E\cap <!--\; \cap \; -->F)$? What is $(\genfrac{}{}{0ex}{}{48}{12,12,12,12})$?

Show any graph G contains an r-partite subgraph H with e(H) $\ge <!--\; \ge \; -->\frac{r-<!--\; -\; -->1}{r}e(G)$
I'm trying to show that for any $r\ge <!--\; \ge \; -->2$ any graph G contains an r-partite subgraph H with e(H)$\ge <!--\; \ge \; -->\frac{r-<!--\; -\; -->1}{r}e(G)$

Proving a binomial sum identity $\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{n}{k})\frac{(-<!--\; -\; -->1{)}^k}{2k+1}=\frac{(2n)!!}{(2n+1)!!}.$

Let's call $S$ the infinite string that is made by concatenating the consecutive positive integers written down in base $10$. Thus,
$S=12345678910111213141516171819202122232425\dots <!--\; \dots \; -->$
Any number in $S$ occurs multiple times. The first occurrence of $3$ is in the third position of the series, the second occurrence is in the seventeenth position, and so on.
How do I find the position of the hundredth occurrence of $3$? Is there a pattern?

In a box there are 16 ice-creams: 6 lemon flavor,4 mint flavor and 6 strawberry flavor.When we extract two ice-creams,what's the probability of getting two different flavors,given that at least one is strawberry flavor.
That's my solution :
$P(A)$ = "Different flavors" = $\frac{(\genfrac{}{}{0ex}{}{16}{2})}{16!}$
$P(B^c)$ = "At least one is strawberry flavor" = $1-<!--\; -\; -->\frac{(\genfrac{}{}{0ex}{}{10}{2})}{16!}$
We want $P(A|B^c)$ using conditional probability, where I go wrong?

Set containing numbers $1$ to $49:\{1,2,3,\cdots <!--\; \cdots \; -->,49\}$ Divide the set into $7$ subsets such that each subset should contain $7$ elements and sum of the elements of each subset should be $175$.
Is it possible to prove that such subsets exist?