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 Apr 7 answered Proof by Induction: Number of bit strings of length $n$ starting with a 1 or ending with a 0 Mar 15 answered Probability to find at least one alphabetically ordered subset of K elements in a set of N elements Mar 15 comment Probability to find at least one alphabetically ordered subset of K elements in a set of N elements OP means any positions. Only that is consistent with 119/120. 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