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

Home > Probability
Never 0 mod 3 (Posted on 2016-03-17) Difficulty: 3 of 5
Let k be a positive integer. Suppose that the integers 1, 2, 3, ...3k, 3k + 1 are written down in random order.

What is the probability that at no time during this process, the sum of the integers that have been written up to that time is divisible by 3?

Source: Putnam competition

No Solution Yet Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
re(4): Solution | Comment 6 of 7 |
(In reply to re(3): Solution by Charlie)

I found it:


 there are (k!*(k+1)!*(3k)!)/((2k)!*k!) = (k+1)!*(3k)!/((2k)!) ways to create a desires sequence out of all (3k+1)! possible sequences.

The first  (k!*(k+1)!*(3k)!)/((2k)!*k!)  uses the combination formula for C(3k,k) for the positioning of the multiples of 3, rather than using permutations.  That means 1346572 and 1643572 count only once in the numerator, but twice in the (3k+1)! denominator of overall permutations of the 7 (in this instance) numbers.  The multiples of 3 are not identical and their order counts just as much in the numerator as in the denominator.

  Posted by Charlie on 2016-03-17 23:21:55
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 (9)
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