Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Questions related to permutations, bijections from a finite (or sometimes infinite) set to itself.
0
votes
1
answer
194
views
Permutations which avoid consecutive entries of the form (m,m+1)
I've just started understanding permutations but could not understand as to how I can come up with a general formula for this problem.
Consider all permutations of the numbers 1 to n. … For a given n, count the number of good permutations.
For example, for n = 3, the good permutations are:
1, 3, 2
2, 1, 3
3, 2, 1
Come up with a form of an answer which can be easily calculated. …