We choose randomly a subset $\mathit S$ of size 25 from the set $\{ 1,2,...,100\}$  , what is the expected value of number of Consecutive numbers in $\mathit S$?    Consecutive numbers in $\mathit S$: is a pair $\{ i,i+1\}$ where i$\in$$\mathit S$ and i+1$\in$$\mathit S$.
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if we choose a size $25$ subset of the set: $\{1,2,....100\}$ what is the expectancy of the number of sequential pairs in the subset? expectancy still confuses me, can anybody help?
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Greetings Community! The above problem you see is a combinatorics problem I could not solve. :( This problem is from ZIO $2010$, P$1$.     Here is what I did: Observe that every graph can be divided into "sub-groups" which may help in solving. We first remove the empty set. Then the singleton set. Consider the bigger but smaller than the sub-parts example:    The singleton sets are $\{4,5,6\}$ which are at the least position in the graph.   The dual sets are $\{2,4\}, \{3,5\}, \{3,6\}$, the bottom - two portions in the graph. The triplet sets are $\{2,4,5\}, \{4,5,6\}$ which we may manually count but for the bigger graphs in the sub-part, this may prove to be difficult and this is where I am stuck.   Any help would be appreciated. Thanks.
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I am trying to solve as many problems as I can but I am a little confused on this one. It is in chapter 1, problem 9 (Probabilistic method, Alon &amp; Spencer). The problem says   Let $F$ be a finite collection of binary string of finite lengths and assume that no two distinct concatenations of two finite sequences of codewords result in the same binary sequence. Let $N_i$ denote the number of string of length $i$ in $F$. Prove   $$\sum_i \frac{N_i}{2^i}\leq 1$$  The problem that I have is that I understand the problem as if $w_1w_2=w_3w_4$ then we must have $w_1=w_3$ and $w_2=w_4$, but if this is what they mean then let $F=\{1,0,100\}$ then the possible contatinations are: $11,00,10,01,1001,1100,0100,1000,100100$, so no two are the same, but then we would have $N_1=2,N_3=1$, so $N_1/2+N_3/2^3=2/2+1/8&gt;1$. I assume I must be understanding the condition incorrectly.
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Q.20 What is the number of ordered pairs $(𝐴, 𝐵)$ where $𝐴$ and $𝐵$ are subsets of $\{1,2, . . . ,5\}$ such that neither $𝐴 ⊆ 𝐵$ nor $𝐵 ⊆ 𝐴$?  Ans: 𝟓𝟕𝟎  Hint: Use principle of Inclusion-Exclusion.  Solution: Let $X$ denote the set of all ordered pairs $(A, B)$ when $A ⊆ B$. Similarly let $Y$ denote set of all ordered pairs $(A,B)$ when $B ⊆ A$. The question asks to find $$𝑛(X'\cap Y') = 𝑛(𝑆) − 𝑛(𝑋\cup𝑌) $$$$= 𝑛(S) − 𝑛(X) − 𝑛(Y) + 𝑛(X\cap Y) $$$$= 2^{10} − 3^5 − 3^5 + 2^5 = 570$$.    How did this happen? What is $S$? If $S={1,2,3,4,5}$, shouldn't $n(S)=2^5$? And where do the $3^5$s come from? I cannot understand this.  Also, are there other ways to solve this?
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Calculate $$A={1\over {n\choose 0}}+ {1\over {n\choose 1}}+ ...+{1\over  {n\choose n}}$$  and  $$B={1\over {n\choose 0}}- {1\over {n\choose 1}}+ ...+{(-1)^n\over  {n\choose n}}$$     My idea for $A$ is some probabilistic reasoning. Color the sets $$\{\}, \{1\}, \{1,2\}, \{1,2,3\},...\{1,2,...,n\}$$ and ask our self what is the probability that I choose colored set among all sets. Clearly this is exactly ${n+1\over 2^n}$ and on the other side it is $A$:  probability that I take empty set is ${1\over {n\choose 0}}$,   probability that I take colored set with 1 element is ${1\over {n\choose 1}}$  probability that I take colored set with 2 elements is ${1\over {n\choose 2}}$  and so on ...   So $A ={n+1\over 2^n}$. But I have no idea how to attack $B$.
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Let's say I have numbers each taken in a set $A$ of $n$ consecutive naturals, I ask myself : how can I found what are all the unique multisets, which could be created with $k$ elements of this set $A$?  For example I've got $A=[1,2,\dots,499]$. If I wanted to create unique multiset of 3 elements, I would search all the multisets $\{a,b,c\}$ such as $a\leq b\leq c$ and then have all the unique multisets. As such for each elements $a$ there is $499-(a-1)=500-a$ possibilities for $b$ and $500-b$ possibilities for $c$. Unfortunately I'm stumped here and can't find the number of possible combination, as I didn't do any maths for years. I know that I should have some kind of product, but I don't know how to find the product anymore.  So first, I would like to know if I am right in my original assumption, or if I am looking in the wrong direction. Second I would like to approach that as if I were doing an homework, as I would like to understand the logic behind it: What would be the formula to give the numbers of multisets of $k$ elements from a set $A$ of $n$ consecutive naturals?  P.S. I'm totally new to math.SE, as such, I'm not sure I tagged the post appropriately.
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Suppose that a set has an odd number of elements. Explain why half of the subsets will have an odd number of elements .  Now assuming set A is the set with an odd number of integer elements{1,2,3,4,5}  Subset b ={1,2}  subset c={3}  Subset d={4,5}  Now there are two subsets with even number of elements and one subset with an odd number of elements.It seems to contradict the earlier theorem that i have to prove.I think that i am missing something.
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Suppose there are $N$ cards, $1,2,\dots,N$.   We start drawing cards (with replacement), until each card has been drawn at least once (we stop when the last card is drawn for the first time).   Let $x$ be the number of cards that have been drawn exactly once.   What is the expected value of $x$?   The answer is supposed to be $1 + \frac{1}{2} +\dots+ \frac{1}{N}$.
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I need help proving that for every positive integer $n$, there exist $n$ consecutive positive integers, each of which is composite. The hint that came with the problem is: Consider the numbers $2+(n+1)!,3+(n+1)!,...,n+(n+1)!,n+1+(n+1)!$.  I honestly have no idea how to begin here. I am specifically confused about what it means when it says "there exist $n$ consecutive positive integers". Does this mean that if $n = 5$, for example, then somewhere in the positive integers there are 5 consecutive composite integers, and that we want to prove that? I get that composite means that they are not prime, so they have more factors than 1 and themselves.  Any help would be greatly appreciated.  Thank you!
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Given a list $A$ of $n$ positive integer numbers. We're gonna play this game:  $1 -$ Take randomly $2$ numbers of $A$. $2 -$ Delete this $2$ elements of $A$. $3 -$ Insert in $A$ their gcd and lcm. $4 -$ Go to step $1.$  Prove that after some quantity of steps, $A$ doesn't change its elements.
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I confirmed on  thread that there are as many as even natural numbers as there are natural numbers.   Question : Suppose I have selected a number $n \in \mathbb N$; what is the probability that $n$ is even?  My Thought : $\text{Probability} = \dfrac{\text{n(E)}}{\text{n(S)}}$ Here $\text{n(S)}$ is the set of all natural numbers i.e. $\mathbb N$, and $\text{n(E)}$ is set of all even natural numbers. Since it is proved that number of elements is the set $\mathbb N$ is exactly the same as the number of elements in the set of natural numbers (it’s very easy to put the set of natural numbers, $\Bbb N=\{0,1,2,3,\dots\}$, into one-to-one correspondence with the set $\text{E}=\{0,2,4,6,\dots\}$ of even natural numbers; the map $\Bbb N\to \text{E}:n\mapsto 2n$ is clearly a bijection.) ; Thus, Probability $= \boxed 1$ I know this is definitely wrong.Probability must be $0.5$. But where am I wrong? Can anyone explain ? Thanks!
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More generally, if a set $S$ has cardinality $\mathfrak{m}$, how many of its subsets have cardinality $\mathfrak{n}$? Clearly there are at least $2^\mathfrak{n}$ such subsets. I don't see how many more though.  Thanks
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Given $n_1 $ number of  $1 $'s,  $n_2 $ number of  $2 $'s,  $n_3 $ number of  $3 $'s,  $n_4 $ number of  $4 $'s.  form a sequence using all these numbers such that two adjacent numbers should not be same.  I have tries lot of things but nothing worked.  can somebody tell me how to solve this problem?
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A subset of $\{1, 2, 3, \dots, 20\}$ containing 10 elements is chosen at random. The largest element of this subset is noted. Find the expected value of the largest element.  I tried to solve this problem by finding the probability of each element being the largest one, but I got confused and would like some help. :)
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Q: How many arrangements of a,a,a,b,b,b,c,c,c are there such that  $\hspace{5mm}$ (i). no three consecutive letters are the same?  $\hspace{5mm}$ (ii). no two consecutive letters are the same?  A:(i). 1314. ${\hspace{5mm}}$ (ii). 174.  I thought of using the General Principle of Inclusion and Exclusion along with letting $p_i$ denote a property that.. , and doing so, I will evaluate $E(0)$, which gives us the number of arrangements without any of the properties.   How do I go about doing that? I am trying to use the method stated above in solving this question, but I am unable to generalize the properties $p_i$.
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What is the next number in the following set ?  $$1,11,21,1211,111221, \ldots$$
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Prove that for each $n$, there exist $n$ consecutive integers, each of which is divisible by a perfect square larger than $1$. Can any one help with this? I was stuck because I know how to solve for a particular $n$ value, but I was helpless in the generalized form.