# Homework Help: I always hate this - combination / permutation

1. Mar 17, 2012

### kenny1999

1. The problem statement, all variables and given/known data

I know what's happening with the problem, but I just hate myself drawing all possible arrangements, I think I am too stupid, I hope better solution.

Problems:

6 different Cookery books and 8 different History books are put on bookshelf. How many possible arrangements if no two cookery books can be put next to each other.

2. Relevant equations

3. The attempt at a solution

My first approach was to think about the number of arrangements, without considering the difference within the same category of books. C for cookery , H for history

then CHCHCHCHCHCHHH
HCHCHCHCHCHCHH

...........

then it comes to a hard time, stupid time, trying to figure out all possiblity by human minds.

It is very easy to make mistakes and very time consuming and hateful....

I think I am too stupid and I don't think it's the solution. Would anyone teach me any other possible solution?

Thanks

2. Mar 17, 2012

### tal444

Hint: You have 6 choices for the first cookbook, then 8 choices for the first history book, then 5 for cookbook, etc... Multiply those together.

3. Mar 17, 2012

### rcgldr

I'm not sure this is the best approach, but note that you can reduce the problem somewhat by noting that there is at least one history book between each cookery book. That uses up 5 of the 8 history books. Then you're left with how many ways are there to distribute the remaining 3 books into the 7 possible positions either outside or between the 6 cookery books:

X C X C X C X C X C X C X

This can be futher reduced if you take into account that 3 books can only occupy 1, 2, or 3 of the 7 possible positions for any specific combination. This will give you all the combinations. If you want the number of permutations, what would you do with the number of combinations that you determined were possible?

Last edited: Mar 17, 2012
4. Mar 18, 2012

### Joffan

Given the assumption you have made at the beginning is that the cookery books are indistinguishable from each other, as are the history books, rcgldr's scheme of seven locations for the additional three books gives a good way forward, akin to throwing three identical marbles into seven numbered buckets.

Once you have that answer, increasing it to reflect the potential for distinguishable cookery and history books is fairly straightforward too.

And seriously, who would have that many identical copies of two such different books and then jumble them up on one shelf? :-)

5. Mar 19, 2012

### kenny1999

hey, i just figure out the answers for the number of possible arrangement for 6 and 8 indistinguisble C and H books

but i am not sure if it's true. my solution:

7 + 7x6 + 7C3

7 for all the three Historic books together
7x6 for any two of the three history books together. 112, 113, 114, 221, 223, 224 etc.........
7C3 for no any of the three historic books together 123, 124, 234, etc.......

Right??

Then I only have to multiply (7+7x6+7C3) by 6! x 8! to get the all possible arrangements for 6 DIFFERENT cook book AND 8 DIFFERENT hisotry books,.... am i right?

Last edited: Mar 19, 2012
6. Mar 19, 2012

### rcgldr

Yes, this seems to be the correct answer, unless I got it wrong also.

For all 3 books into 1 position, it's 7C1 = 7.

For 3 books into 2 positions, there are two combinations, 1st postion has 1 book, 2nd position has 2 books, or 1st position has 2 books, and 2nd positon has 1 book. There's a formula for this, r (identical) books into n positions, no position empty = (r-1)C(n-1). So this becomes 2C1 x 7C2 = 2 x 21 = 42 (= 7 x 6 when counted by the method you used).

You already figured out that for 3 books into 3 positions it's 7C3.

Last edited: Mar 19, 2012
7. Mar 20, 2012

### Joffan

Just as an additional check, putting $x$ identical items into $y$ distinct containers/categories is actually given by $(x+y-1)Cx$, or in this case $9C3$, which agrees with your case-by-case assessment.

(Why? because it is equivalent to distinct arrangements of a set consisting of the $x$ items and the $(y-1)$ dividers between categories).

8. Mar 21, 2012

### azizlwl

I'm too new to permutation and let me try.

The easiest of the permutation if equal number books are given.
The permutation for 6 books for each subject can be given as 6!.6!
Now create 6 history books, 4+(2)+(2) or 5+(3)
CHCHCHCH Total ways= 6!.6!.8C4 =A+ 6!.6!.8C3=B

Now HCHCHCHC OR HCHCHCHCH 6 ways or 7 ways.
6 ways as A+B
7ways= 6+(2) or 5+(1)+(2)=6!7!.8C2=B or 6!.7!.8C3=C

Total ways=2(A+B)+C

Last edited: Mar 21, 2012