# Another Permutation question

• Sep 2nd 2011, 06:22 AM
richardyuen2144
Another Permutation question
Here is my question
In how many ways can 8 library books be arranged on a shelf if 3 of the books cannot be all together?

What i did is:
3 books cannot be together meaning there can be two, one or even none of the three 'special' books are in the arrangements.

If there are two 'special' books:
7 books available for arrangement: 7P7 = 5040
And because 2 out of 3 books are available so there will be 6 possible arrangements for the 'special' books. Therefore, 7P7 x 6 = 30240

If there are one 'special' book:
6 books are available for arrangement: 6P6 = 720
1 out of 3 'special' books are available so 3 possible arrangement for the 'special' books. Therefore, 6P6 x 3 = 2160

If there no 'special' book:
5 books are available for arrangement: 5P5= 120

Adding them all together should get 30240 + 2160 + 120 = 32520 arrangements

What did i do wrong?(Worried)(Worried)
• Sep 2nd 2011, 06:31 AM
Plato
Re: Another Permutation question
Quote:

Originally Posted by richardyuen2144
Here is my question
In how many ways can 8 library books be arranged on a shelf if 3 of the books cannot be all together?

There are five books that are separators.
__X__X__X__X__X__ which create six places to place the three special books.

BTW this is were all three are separated from one another.