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Dec
20
suggested rejected edit on Upper bound for a number of subsets of $\{1, \dots, n\}$
Dec
20
revised Number of strings lenght $n$ with no consecutive zeros
Corrected sign of $k$ as suggest by Deepak
Dec
20
comment Number of strings lenght $n$ with no consecutive zeros
@DeepakGupta You are right, I got the sign of $k$ wrong. Please see the edit.
Dec
18
awarded  Yearling
Dec
15
answered Upper bound for a number of subsets of $\{1, \dots, n\}$
Dec
4
answered Number of strings lenght $n$ with no consecutive zeros
Sep
12
comment N gunmen in a field
If we retain all other conditions except for the fact that shooters are in 3(or $n$ dimensional euclidean space), (a) will hold. What about (b) and (c)?
Sep
12
comment Is my assumption about dependencies for this particular setup correct?
equivalent to $\mid T \mid$
Sep
11
answered Is my assumption about dependencies for this particular setup correct?
Sep
7
answered N balls having M different colors in a box, how many times do I need to pick to get one particular color?
Sep
2
answered Number of divisors $d$ of $n^2$ so that $d\nmid n$ and $d>n$
Aug
26
revised Average length of a cycle in a n-permutation
removed clarification after the question was reopened
Aug
26
comment Average length of a cycle in a n-permutation
Proof of the fact: The expected number of k-cycles in a random permutation of $[n]$ is $\frac{1}{k}$. Let $x_i=1$ if $i$ is a part of a k-cycle, else 0. $\frac{\sum_{i=1}^{n}x_i}{k}$ counts the number of k-cycles of $[n]$. We have $E \left( \frac{\sum_{i=1}^{n}x_i}{k} \right)=\frac{1}{k} \sum_{i=1}^{n}E(x_i)=\frac{1}{k} \sum_{i=1}^{n}\frac{1}{n}=\frac{1}{k}$ Proof of the fact: Probability that 1 belongs to a k-cycle is 1/n (independent of k). Apart from one, a k-cycle can be constructed in $\binom{n-1}{k-1}(k-1)!(n-k)!$ . Dividing by $n!$ gives $\frac{1}{n}$
Aug
26
revised Average length of a cycle in a n-permutation
clarification about why this question is not a duplicate
Aug
26
revised Average length of a cycle in a n-permutation
clarification about why this question is not a duplicate
Aug
26
comment Average length of a cycle in a n-permutation
@brian-m-scott : It would be great to hear about the approach you had on mind.
Aug
26
accepted Average length of a cycle in a n-permutation
Aug
26
comment Average length of a cycle in a n-permutation
Thanks Brian! Please let me know if my further analysis is incorrect. Total number of k-cycles (in all permutations) is $n!/k$ and total number of cycles (in all permutations) is $n! \cdot H_n$. Hence, $p(length=k \mid cycle)$ (probability that a random cycle has length k) is $\frac{n!/k}{n! \cdot H_n}=\frac{1}{k \cdot H_n}$. From here, avg cycle length=$\sum_{k=1}^{n}k\cdot p(length=k | cycle)=\sum_{k=1}^{n}k\cdot \frac{1}{k\cdot H_n}=\frac{n}{H_n}$
Aug
26
asked Average length of a cycle in a n-permutation
Jan
13
answered Number of upward closed subsets