All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers > Sequences
Events (Posted on 2005-02-10) Difficulty: 1 of 5
Suppose there were three events that could be done in any order. Let these events be A, B and C. There are a total of 6 possible ways the events can be performed.

However, this time, there is a restriction. B can only be performed after A. In other words, A must be performed before B can, and ACB is also accepted.

Your task is to find a general formula for X number of events, and Y number of restrictions.

None of the events is mentioned in more than one restriction.

No Solution Yet Submitted by wcaexqdz    
Rating: 2.0000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution idea - plz read and see wot u think - if youve already got the answer | Comment 4 of 14 |

The relationship between events and restrictions is easy! But it involves a new variable.

If you call 3 events term 1, you can call 4 events (and the corresponding restrictions) term 2, and so on and so on... 

You multiply the restrictions by 2, and if the term number is odd, you add on 1 after, to get the events.

2y+1 (if term is odd) = x

It's not that general, but general enough, I think. One question though - by restriction, do you mean strictly one event must come before/after another event, or do you mean things like one event must be first, one last, etc. Or is the whole problem much more complicated???  


  Posted by Angela on 2005-02-15 19:10:59
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information