Combinatorics: Examples, Equations, and Practice Problems

Suppose $S\subset <!--\; \subset \; -->\mathbb{Z}$ (set of integers) such that
1) $|S|=15$
2) $\mathrm{\forall <!--\; \forall \; -->}s\in <!--\; \in \; -->S,\mathrm{\exists <!--\; \exists \; -->}a,b\in <!--\; \in \; -->S$ such that $s=a+b$
Show that for every such $S$, there is a non-empty subset $T$ of $S$ such that the sum of elements of $T$ is zero and $|T|\le <!--\; \le \; -->7$.

There are 4 rows and 3 columns in a table, and each slot is painted with black or white with equal probability, and I wish to find the probability that one, and only one of the rows is painted black. What I did was:
$P=4\cdot <!--\; \cdot \; -->(1/2{)}^3\ast <!--\; \ast \; -->(7/8{)}^3=\frac{343}{1024}$
and it indeed seems to be the correct answer. I thought I should also try solving it using combinatorics, but that is something I fail to do. My try is:
$\frac{4\cdot <!--\; \cdot \; -->1\cdot <!--\; \cdot \; -->(2^9-<!--\; -\; -->7)}{2^{12}}=\frac{505}{1024}\ne <!--\; \ne \; -->\frac{343}{1024}$
what have I done wrong?

There are 9 boys and 7 girls sitting around a table. If a boy sits next to a girl we call it a switch. Example: In BBBBBGGGBBGBGGG there are 6 switches.(If a boy sits next to 2 girls, there are 2 switches counted.) What is the expected number of switches?

Tried playing with the formula of Conditional probability and some variations of combinatorics, because i know for sure that the 6th toss was head (we stop on the 3rd time we get head, and it's given to us that we stopped after 6 tosses). I know the answer is 0.4, but I cant unserstand how to get this answer, any help?

There are three colors of counter, red, green and blue.
There are 5 red, 3 green and 2 blue counters already in the bag.
Another counter is added to the bag (completely randomly).
A counter is chosen from the bag.
Given that the counter chosen is green, what's the probability that the counter added to the bag was green?

Prove that $(\genfrac{}{}{0ex}{}{2n}{n})=\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}{(\genfrac{}{}{0ex}{}{n}{k})}^2$
So far I have tried writing the right hand side in different ways: expressing it in its factorial form and have tried to implement the identity
$(\genfrac{}{}{0ex}{}{n}{k})=(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->1}{k-<!--\; -\; -->1})+(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->1}{k})$
but have not gained any new ground. If any one has an algebraic proof, or even a simple combinatorics proof that is intuitive and used with an example that would be preferable.

What is the probability that 2 or more numbers out of 4 are among the drawn lottery numbers (6 out of 49 without replacement)?

For any even $n$, say $n=2m$, a row complete Latin square of order $n$ can be formed by writing down
$0,1,2m-<!--\; -\; -->1,2,2m-<!--\; -\; -->2,3,\dots <!--\; \dots \; -->,m+1,m$
as the first row and then developing subsequent rows by adding $1$ modulo $n$.
I'm not quite clear on how that goes, such as what the difference is between $2m$ and $m$. $2m$ stands for $n$, does $m$ stand for modulo? I'd like to see and example with an even number or two, such as $4$, and $6$.
I tried, $2m=4$
$0,1,3,2,2\dots <!--\; \dots \; -->$
wait, that can't be right.

Let's say, I have 4 yellow and 5 blue balls. How do I calculate in how many different orders I can place them? And what if I also have 3 red balls?

How can we show that the number of pairs $(a,b)$ (where the pairs $(a,b)$ and $(b,a)$ are considered same) with $\mathrm{lcm}<!--\; \; -->(a,b)=n$ is equal to
${\displaystyle \frac{(2e_1+1)(2e_2+1)...(2e_k+1)+1}{2},}$
Where
$n=p_1^{e_1}{p}_2^{e_2}\dots <!--\; \dots \; -->p_k^{e_k}$, pis are prime for all $1\le <!--\; \le \; -->i\le <!--\; \le \; -->k$.

What is the probability that a bridge hand will contain 13 cards of the same suit?

How many ways are there to color faces of a cube with N colors if two colorings are the same if it's possible to rotate the cube such that one coloring goes to another?

Consider a deck of 52 cards (there are 13 cards of each suit).
a. How many different ways can you arrange, in a line, the 13 cards of the suit of diamonds, so that the cards 7,8,9 and 10 stay together either in the beginning or the end of the line?
b. 8 cards are extracted from the deck. What is the probability there is an ace and at least 3 queens in the extracted cards?
Is there any method to solve probability questions?
What strategies/lines of though do you use when tackling this kind of problems?
Could you solve these problems using a method or several strategies and describe them?
Is there any single method you can use to solve both of these seemingly different problems?

Define a matching problem to be the following:
1) n numbered boxes and n numbered cards
2) you have to place all the cards. No empty boxes and no left behind cards are allowed.
So you end up with a permutation of length n. The number of possible permutations being n!
Let X be the variable of x matching places in an ordering.
Only 1 permutation is correct. So $P(X=n)=\frac{1}{n!}$
I would like to calculate $P(X=2)$, $P(X=3)$ and so on.
$P(X=1)=0$ obviously...

Lets assume we have a box with 45 coupons labeled 1-45. Now in this case I would like to adjust the CCP such that I can calculate the expected value (amount of draws necessary) to collect 10 specific items. For example item 1-10. How do I adjust my model such that I can calculate the amount of draws necessary to collect each item n times.

Suppose that I am buying cakes for a party. There are $k$ different types and I intend to buy a total of $n$ cakes. How many different combinations of cakes could I possibly bring to the party?

A manager must form a team of 5 from among 5 employees from group A and 6 employees from group B. If all employees have an equal chance of being selected, what is the approximate probability that a randomly selected group of 5 will consist of 2 employees from group A and 3 employees from group B?

With a standard 52 deck (13 types and 4 suites), what is the probability that a poker hand contains cards of five different types? A poker hand is a set of 5 cards.
The obvious answer is A / B where A is the number of poker hands that have 5 different types of cards and B is the total number of poker hands which is C(52,5).
Right now I think A = C(13,5) since there are 13 different types of cards I choose 5 from them but I am unsure if A = 4 * C(13,5) instead, where 4 is the different suites.

We have at our disposal the following: the three letters a,b,c , and the five digits 1,2,3,4,5 . We have to form with them all the possible passwords of six (6) characters, under the condition that there must be at least one letter and at least one number in each password. Other than this there are no more restrictions (and, thus, one can repeat at will numbers, letters and etc.)

Evaluate $\underset{k\ge <!--\; \ge \; -->0}{\sum <!--\; \sum \; -->}(-<!--\; -\; -->1{)}^k{(\genfrac{}{}{0ex}{}{n}{k})}^2$ using sign reversing involutions. When $n$ is odd, the problem is trivial : let $[n]=\{1,2,\dots <!--\; \dots \; -->,n\}$ consider all pairs of subsets $\{(A,B)\in <!--\; \in \; -->2^{[n]}\times <!--\; \times \; -->2^{[n]}:|A|=|B|\}$, and $(A,B)$ has sign $(-<!--\; -\; -->1{)}^{|A|}$ and the sign-reversing involution is $(A,B)\to <!--\; \to \; -->([n]-<!--\; -\; -->A,[n]-<!--\; -\; -->B)$. Any hints on how to approach this problem for even $n$ will be appreciated.