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August 6th, 2007, 08:45 PM   #1
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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?
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August 7th, 2007, 05:18 AM   #2
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Well, there are 10! ways to put the books, and then you just have to arrange the indistinguishable bookshelves amongst the 8 possible positions.
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August 7th, 2007, 07:11 AM   #3
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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!
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August 8th, 2007, 08:57 AM   #4
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I notice you state that the books are different, but are the shelves different?
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August 8th, 2007, 09:40 AM   #5
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If the shelves are indistinguishable, then the answer is 6*10!.
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August 8th, 2007, 10:58 AM   #6
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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
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August 8th, 2007, 11:00 AM   #7
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kitaplık=shelves
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August 8th, 2007, 12:51 PM   #8
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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+k-1)!]/[n!(k-1)!]=(n+k-1)C(k-1).

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.
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August 8th, 2007, 01:39 PM   #9
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ı like long solutions yes you right we have formula c(n+r-1,r-1)

but ı dont like this
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August 8th, 2007, 02:01 PM   #10
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Quote:
Originally Posted by roadnottaken
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+k-1)!]/[n!(k-1)!]=(n+k-1)C(k-1).

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.
Well-phrased. I gave an outline of that method, but your post is much clearer.
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