1. ## question on permutation

Two Mathematics books, three Chemistry books and four Physics books are to be arranged in a row on a shelf. Given that all the books are different, find the number of ways of arranging the books if no two Physics books are placed together.

I am able to find out the number of possible arrangement for the 4 physics books so that no two of them are placed together, which is 15.
so 15x4!x5!=43200.
But this method is a bit stupid since I need to count the possible arrangement one by one.

2. Don't post the same question to multople areas on the MathHelpForum.

It confuses the helper and makes work for the Moderators.

RonL

3. ooops...sorry. I post it at the wrong section in the first place

4. Hello, acc100jt!

I can't locate your duplicate post, so this may be a repeat of someone else's solution.

Two Math books, 3 Chemistry books and 4 Physics books are to be arranged in a row on a shelf.
Given that all the books are different, find the number of ways of arranging the books
if no two Physics books are placed together.

Consider the 2 Math books and 3 Chemistry books placed in a row
. . with a space between them and at the ends.

. . $\displaystyle \_\_\:X\;\_\_\;X\;\_\_\;X\;\_\_\;X \;\_\_\;X\;\_\_$

These five books can be arranged in $\displaystyle 5! = 120$ ways.

Now place the 4 Physics books.
We have a choice of six spaces and we will choose four of them.
There are: .$\displaystyle P(6,4) = \frac{6!}{2!} = 360$ ways.

Therefore, there are: .$\displaystyle 120 \times 360 \:=\:43,200$ possible arrangements.

5. Thanks a lot!! soroban!!!!