So this is my first time encountering this type of problem, and I'm at quite a beginner level. If you could help me get on the right track I'd appreciate it.  Now if i'm picking r objects from n objects that's going to be  $n\choose r$   now there would be $\frac {n!}{(n-r)! r!}$ combinations (i think)  if we just wanted  to know ways where each object appears an odd number of times, wouldn't that be half of the total possibilities?   My logic and math may be way off here, but I'm just a beginner. I feel like this makes sense because all the odd possibilities would be half the amount of all possible possibilities right?  So if you were to pick r objects from n objects, how many ways are there to pick so that each r object appears an odd number of times. Is what the question is asking
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Let $A$ be a non-empty set and $n$ be the number of elements in $A$, i.e. $n:=|A|$.  I know that the number of elements of the power set of $A$ is $2^n$, i.e.  $|\mathcal{P}(A)|=2^n$.  I came across the fact that exactly half of the elements of $\mathcal{P}(A)$ contain an odd number of elements, and half of them an even number of elements.  Can someone prove this? Or hint at a proof?
6646
Background: I'm working on  for , and I've run into a problem with how I'm creating the individual matchups (groups of 4 entries).  Currently, I'm simply generating all combinations of 4 elements (in the combinatoric sense) of entries, but that gets really big really quickly:  $$x \choose 4$$    (from )  Therefore, my question is: How can I choose subsets (of length $n$) of a set such that each element of that set appears in the same subset as every other element at least once, while maintaining a minimal amount of subsets?  I can figure this out for $n = 2$, since the solution is quite intuitive: simply combine the first element with every other element, combine the second element with every other element that comes after it, etc. For a set of length $L$, this yields:  $$\sum_{x=1}^{L-1}(L - x)$$    (from )  which is much smaller and more manageable.  However, I still have not been able to figure out: How can I generalize this to any $n$?
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I perfectly understand the permutations formula i.e. if you have $n$ things how many ways can you rearrange it if taken $k$ at a time (or if you have $k$ slots)? So you draw the following tree. And the formula comes out naturally. $$^nP_k = \frac{n!}{(n-k)!}$$  I get that it is including all of the duplicates, too. And to get rid of them we use the combinations formula: $$^nC_k = \frac{n!}{k!(n-k)!}$$ Can somebody explain why we have to divide by the &quot;number of ways to re-arrange the slots&quot; to get rid of the duplicates? I understand that it works and the division is used to scale a number down but I am not getting that &quot;aha moment&quot; of why we do it. Can somebody explain it?
32480
Say, we have 4 red balls, 4 blue balls, 4 green balls, 4 yellow balls. How many ways are there to form a sequence of 10 balls such that every color of ball occurs at least twice?  My thought: r = red b = blue g = green y = yellow * = undetermined  Every arrangments should in this form: rrbbggyy**  Case 1: two * are same  $\frac{\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}10!}{4!2!2!2!}$  Case 2: two * are different  $\frac{\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}\binom{4}{2}10!}{3!3!2!2!}$  Result = Case 1 + Case 2  Any suggestions would be appreciated!
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Suppose 27 cubes are stacked together, suspended in the air, to form a larger cube.  The cube is then painted on all the exposed surfaces and dried.  The smaller cubes are then randomly permuted in both spatial position and spatial orientation to form another large cube.  What is the probability that new larger cube is identical to the original?  This is not homework, it was passed on by a friend who found it somewhere on the internet, and they do not remember where.
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I don’t know how to handle this 3 players case, may anyone give a hand? Thanks!     When playing card game, there are 3 players playing 32 cards, each   receive 10 cards, 2 remind on the table. Overall are 4 aces and 4   kings in the game. How likely is P (E | D)? D = "The table remains 2   aces" E = "I (at least) get one king".   I guess I should have steps as follows.   P(E) = P(1King)+P(2K)+P(3K)+P(4K) P(D) = (4/32) * (4/32) = 1/256 P (E | D) = P(E) + P(D)   Equations may be used:  Counting combinations      The number of combinations of n objects taken r at a time is     nCr = n(n - 1)(n - 2) . . . (n - r + 1)/r! = n! / r!(n - r)!                 Number of favorable outcomes P(I) =     Total number of possible outcome
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Number of ways to select k people from a round table of n people so that no two selected people sit next to each other.    I tried the followings.  k = 1, Number of ways = n  k = 2, Number of ways = n * (n-3)  k = 3, for the first person, we have n choices        for the second person: Case 1: we can select the person sitting next next to the first person, so we have 2 choices for the second person, and the third person we have n-5 choices. Case 2: we select the remaining people other than those in case 1, so we have n-5 choices for the second person, and the third person we have n-6 choices. Then, the number of ways = Case 1 + Case 2 = n*2*(n-5) + n*(n-5)*(n-6)  For k >= 4, it seems that there are many cases for each step, so is there any better way to generalize the formula to select k people?
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Given a set of whole numbers of size $n$, how many unique combinations of these whole numbers are there?   For example, for the set $\{1,2,3\}$ the unique combinations are   $\{1,2,3\}, \{1,2\}, \{1\}, \{2\}, \{3\}, \{2,3\}, \{1,3\}$   The number of combinations for $\{1,2,3\}$ is $7$.  Does this generalize to the number of unique combinations in a set is $2^n-1$ where $n$ is the number of elements in the set?
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Before reading the Wikipedia article, my idea to calculate the probability was as follows:  $1-\left(\frac{364}{365}\right)^{_nC_{2}}$  Basically, I thought to compare all combinations of pairs ($_nC_{2}$) and require them to have different birthday ($\frac{364}{365}$). It gave a very similar result to the wiki method, but is slightly different and I can't tell how  my way of thinking is wrong.  Can anyone tell me what I'm missing?
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I know that the partition of $2n$ items into $n$ pairs has something to do with double factorial, but I am not sure how many possibilities we exactly have.   We can choose such a partition into pairs by first choosing $n$ elements to be in one half and the remaining $n$ in the other half, which gives ${{2n} \choose n}$ possibilities. Then we can take a permutation of both sets and match them accordingly.   This gives $${{2n} \choose n} n! n!=\frac{(2n)! n! n!}{n! n!}=(2n)!$$ possibilities. But with this we overcount, since we do not want the pairs to be ordered. Each of the $n$ pairs is counted twice, wherefore we have to divide by 2 for every pair, hence $\frac{(2n)!}{2^n}$?   Well, another possiblity to do that is just to iteratively pick $2$ elements among the remaining ones, which gives ${{2n} \choose 2} {{2n-2} \choose 2} \dotsb {4 \choose 2}$  I am not sure, but I do not think that this is the same and also, I do not see the connection to double factorial. Would be very glad if you could help me with these 3 possiblities.   Thank you.
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I have never done this type of question (with arbitrary variables in a counting problem), so I had a lot of difficulty. Please let me know if my approach is wrong. Thanks!   "Computer number 1 is assigned no jobs"; "Computers number 1 and 2 are assigned no jobs"; "exactly two of the computers are assigned no jobs" "No computer is assigned more than one job".   Here are my attempts:  Firstly, the total number of combinations is $r^n$, since for each of the n computers, an unlimited number of tasks from r can be allocated.    Since Computer 1 can be assigned no jobs, there are $r^{n-1}$ ways to assign the tasks to the remaining tasks. Thus P = $\frac{r^{n-1}}{r^n}$. Same idea to the one above, P = $\frac{r^{n-2}}{r^n}$ There are $r^{n-2}$ resources to allocate, but since any two computers, there would be $n \choose 2$ different ways to choose the two computers. Thus $P=\frac{(r^{n-2}) {n \choose 2}}{r^n}$ This one I have no idea how to even begin. Please help!
88055
Assuming "complete randomness" and given a string with a length of 20 characters where each character may be one of 62 possible characters:    What are the total number of combinations possible? (Guessing 20 to the power of 62.)  Also, if new strings are randomly selected one after another and added to a list of strings selected so far, how many strings must be selected before the chance of selecting a string that has already been selected is below 1-in-100000 ($10^{-5}$)?   Note: 62 comes from: numeric digits (0-9), uppercase letters (A-Z), and lowercase letters (a-z).
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Let n be an even integer. How many different ways can I choose subsets of the set {1,2...,n} if they all have to have odd size, but the intersection of any two of them has to have even size?  Please help me answer this question or lead me in the right direction. Thank you :)
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I'm reading an algorithm book. In which it mentioned:     Imagine (once again) you have n people lined up to see a movie, but   there are only k places left in  the theater. How many possible   subsets of size k could possibly get in? That’s exactly $C(n,k)$, of   course,  and the metaphor may do some work for us here. We already   know that we have $n!$ possible orderings of  the entire line. What if   we just count all these possibilities and let in the first $k$? The only   problem then is  that we’ve counted the subsets too many times. A   certain group of k friends could stand at the head of  the line in a   lot of the permutations; in fact, we could allow these friends to   stand in any of their $k!$  possible permutations, and the remainder of   the line could stand in any of their $(n–k)!$ possible  permutations   without affecting who’s getting in. Aaaand this gives us the answer!       $$C(n,k)= \frac{n!}{k!(n −k)!}$$      This formula just counts all possible permutations of the line (n!),   and divides by the number of  times we count each “winning subset,” as   explained.   I'm not quite understand why the reminder can in any $(n-k)!$ permutation, if he stands at the end of the queue, won't he let all these people in?  And why I should use $n!$ to divide by $k!(n-k)!$ ? Didn't see that consequence from the metaphor.  If you have better ideas to explain this formula, you can use them of course.
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A coin is tossed $n$ times. What is the probability of getting odd number of heads?   I started this chapter sometimes ago and faced in front of a tough problem. At first I started considering cases. Case-I : The probability of getting 1 head.Case-II : The probability of getting 3 head and so on. But there are many cases. So how can I solved this . Please help me. Thank you!
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If I want to find how many possible ways there are to choose k out of n elements I know you can use the simple formula below:  $$ \binom{n}{k} = \frac{n! }{ k!(n-k)! } .$$  What if I want to go the other way around though?  That is, I know I want to have $X$ possible combinations, and I want to find all the various pairs of $n$ and $k$ that will give me that number of combinations.  For example, if the number of combinations I want is $3$, I want a formula/method to find that all the pairs that will result in that number of combinations are $(3,1)$ and $(3,2)$  I know I could test all the possible pairs, but this would be impractical for large numbers.  But perhaps there's no easier way of doing this then the brute force approach?
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Hello everybody! The problem you see above is a combinatorial problem that I could not solve. :(  Since we are to choose distinct numbers from $\{1,2,3 \ldots K\}$, the numbers must be in increasing order because they are all positive. I think that the difficulty of this problem is in the fact that the Special Sequence can be of any length. We can go about choosing say $n$ elements in $\binom{K}{n}$ ways but again how do we control that? Talking about a recursive solution, I think that is the way to go but I can't get how to do that (that is get a recursive formula for that).  Any help in solving this problem would be much appreciated. Thanks.