# question on permutation

• October 26th 2006, 11:38 PM
acc100jt
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.

• October 27th 2006, 12:33 AM
CaptainBlack
Don't post the same question to multople areas on the MathHelpForum.

It confuses the helper and makes work for the Moderators.

RonL
• October 27th 2006, 12:35 AM
acc100jt
ooops...sorry. I post it at the wrong section in the first place
• October 27th 2006, 09:06 AM
Soroban
Hello, acc100jt!

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

Quote:

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.

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

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

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

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

• October 27th 2006, 05:47 PM
acc100jt
Thanks a lot!! soroban!!!!