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

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.