Permutation Problem

[youngone]

Recently, I was looking through my sister's Maths Textbooks and I saw a very interesting topic I could learn. So for the next few days I learnt the basics of permutations and their applications. All was well until I stumbled upon a question that I was unable to solve. With no one else to turn to (my sister has not yet been taught permutations yet), I decided to post my question on the boards.

 

The question goes like this:

 

There are 9 books arranged on a shelf. 2 of them are labelled A, 3 of them are labelled B and 4 of them are labelled C. How many permutations of the books are there if:

 

a)none of the books labelled A were next to each other.

 

b)none of the books labelled B were next to each other.

 

c)none of the books labelled C were next to each other.

 

I managed to answer question a) but not questions b) and c). I answered question a) like this:

 

Since we want to find the number of permutations with none of books A next to each other, we just need to find number of permutations with books A next to each other and subtract it from the number of permutations without restrictions.

 

So the number of permutations without restrictions is 9!=362880. Next we group the 2 books labelled A and consider it as one block. So, the number of permutations within books is 2!=2 while the number of permutations with the remaining books and the block of books A is 8!=40320. Therefore the number of permutations with books A next to each other is 2!*8!=80640.

 

Finally we just subtract the number of permutations with books A next to each other from the number of permutations without restrictions which is

 

9!-(2!*8!)=282240

 

and it matches the answer at the back of the textbook.

 

I tried to use the same method for the remaining questions but it did not work. You cannot group up three or four of the books together and try and find the number of permutations because you would not take into account the fact that even only 2 books can be next to each other. I also tried grouping up only 2 of the books and using the nPr notation (3P2 or 4P2) but I still got the wrong answer.

 

So, can anyone here help me to solve this problem?

[matt grime]

you need the inclusion-exclusion principle:

 

you did the number of permutations, then subtracted the number with two books together, now add back the number with three books together (and subtract that with 4 books together)

[youngone]

Okay, so I got the answer for the second question which is 9!-(3P2*8!)+(3P3*7!)=151200. But when I tried to get the answer for the next question I keep on getting it wrong. Can you please correct my working? My working is 9!-(4P2*8!)+(4P3*7!)-(4P4*6!).