
Math Events Math Events, Competitions, Meetups  Local, Regional, State, National, International 
 LinkBack  Thread Tools  Display Modes 
August 6th, 2007, 07:45 PM  #1 
Member Joined: Aug 2007 From: turkey Posts: 57 Thanks: 0  Easy problem in turkey math olympiad problem
In how many ways can 10 different books be arranged on 3 shelves if no shelf is left empty?

August 7th, 2007, 04:18 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
Well, there are 10! ways to put the books, and then you just have to arrange the indistinguishable bookshelves amongst the 8 possible positions. 
August 7th, 2007, 06:11 AM  #3 
Member Joined: Aug 2007 From: turkey Posts: 57 Thanks: 0 
ıts very easy the books 10! arranged later the shelfs are x,y,z x+y+z=10 fınd all x;y,z=36 but we find in z+ final answer is 36.10! 
August 8th, 2007, 07:57 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,724 Thanks: 1807 
I notice you state that the books are different, but are the shelves different?

August 8th, 2007, 08:40 AM  #5 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
If the shelves are indistinguishable, then the answer is 6*10!.

August 8th, 2007, 09:58 AM  #6 
Member Joined: Aug 2007 From: turkey Posts: 57 Thanks: 0 
this is my solutions but turkish x+y+z=10 (x,y,,z) kitaplık üçtane 0 almıcaz bos istemiyor çünkü 10 yapan üç sayı arıcağız 1)1,1,8 toplamı 10 yapar bu sayıyı yerini değiştirerek kaç sayı olduna bakçaz 3!/2!.1!=3 sayı 2)2,1,7 3!/1!.1!.1!=6 3)3,1,6 3!/1=6 4)4,1,5 3!/1=6 5)2,5,3 3!/1=6 6)2,2,6 3!/2!=3 7)3,3,4 3!/2!=3 4,4,2 3!/2!=3 çıkar hepsini topladık 36 eder işte bu uzun yolu formulsuzu kitapalrda 10! sekilde dizilir toplam 10!.36 olur 
August 8th, 2007, 10:00 AM  #7 
Member Joined: Aug 2007 From: turkey Posts: 57 Thanks: 0 
kitaplık=shelves

August 8th, 2007, 11:51 AM  #8 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
That's an awful lot of casework. It's better to use the balls and urns argument. If you have n indistinguishable balls which you want to place in k distinguishable urns, then you can represent each ball as a "", and a divider between urns as a "". Now you will have a string of n 's and k 's, such as . Do you see how this is equivalent? Now, the number of orderings of such a string is [(n+k1)!]/[n!(k1)!]=(n+k1)C(k1). This is a lot easier than breaking your argument into cases based on the number of books on each shelf. Because we want each shelf to have at least one book, we can assign one book to each shelf, and then count the number of possible orderings of the other books. This gives us 36, but we must multiply by 10! because the books are all distinguishable. 
August 8th, 2007, 12:39 PM  #9 
Member Joined: Aug 2007 From: turkey Posts: 57 Thanks: 0 
ı like long solutions yes you right we have formula c(n+r1,r1) but ı dont like this 
August 8th, 2007, 01:01 PM  #10  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
 

Tags 
easy, math, olympiad, problem, turkey 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
A practice problem for potential Olympiad students  Turk  Math Events  1  September 20th, 2012 02:09 AM 
Trigonometry olympiad problem.  Math Events  17  March 28th, 2012 07:05 PM  
easy math problem verification  rdnckengineer  Elementary Math  2  February 1st, 2011 10:14 AM 
A problem from 32nd International Mathematical Olympiad  johnny  Math Events  1  September 2nd, 2009 07:55 AM 
A problem from 31st International Mathematical Olympiad  johnny  Math Events  4  September 2nd, 2009 06:06 AM 