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
But the answer is 36000
What did i do wrong?(Worried)(Worried)
Re: Another Permutation question
There are five books that are separators.
Originally Posted by richardyuen2144
__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.