# Thread: Combinaton of Books

1. ## Combinaton of Books

Here's the problem:
There are 12 books on a shelf. If four of them are blue, how many combinations have all four books together? (then it asks to explain)

Here's what I did:

Ok, so there's 12 books, that's 12!. But then, there are two sets: one for blue books and one for the others. So the set for the blue books is 4! and the other is (12-4)! or 8!. So I ended up with this:

$\displaystyle 12!/(4!8!)= 495$

Is this right? I have no way of checking. Also any tips on ways to think of and attack these problems would be nice.

Thanks!

2. You want the number of combinations with the four blue books together. Choosing such a combination is equivalent to choosing:
- first, the 4 "blue locations" (there are 9 possibilities)
- then the order of the books in these "blue locations" (there are 4! possibilities)
- finally the order of the other books, in the other locations (there are 8! possibilities).
So you end up with $\displaystyle 9\times 4!\times 8!$ combinations.

The "tip" here (and in such cases) was to split your problem into choices of locations and then choices of the ordering.

--
What you have computed is $\displaystyle {12\choose 4}=\frac{12!}{4!8!}$, that is to say the number of ways to choose 4 elements in a set of 12. Indeed, there are 12! combinations of books, and by dividing by 4! you get the number of combinations of books when you group together (and count as 1) the combinations where the 4 blue books globally lie on the same positions (they come by 4!). Then, dividing by 4!8!, you get the number of combinations of 12 books, 4 blue and 8 other than blue, when you group together the combinations where the positions of the colors are the same. So you get the number of ways to choose 4 "blue" positions among 12 slots (no matter in which order).

So when you divide, it means you group together (or "identify") certain combinations, which was not the case here.

Laurent.

3. Thanks!

I figured that number was too small.

I do have one question though. How did you come up with the 9?

I am trying to figure out how to relate this to set notation, the multiplication principle, Cartesian product, etc.--which we just went over in class...

Thanks again!

4. Originally Posted by NoFace
How did you come up with the 9?
If the four blue books are together, then either they are the four first(1 2 3 4), or the four after the second position (2 3 4 5), etc., or they are the four last, that it to say the four after the ninth position (9 10 11 12). Look at the position of the left-most book: it goes from 1 to 9, so there are 9 such configurations.

5. That clears it up, thanks!

6. Originally Posted by NoFace
Here's the problem:
There are 12 books on a shelf. If four of them are blue, how many combinations have all four books together? (then it asks to explain)

Here's what I did:

Ok, so there's 12 books, that's 12!. But then, there are two sets: one for blue books and one for the others. So the set for the blue books is 4! and the other is (12-4)! or 8!. So I ended up with this:

$\displaystyle 12!/(4!8!)= 495$

Is this right? I have no way of checking. Also any tips on ways to think of and attack these problems would be nice.

Thanks!
Since 4 books have to be togethet, they make up one unit. Now rest of 8 books plus this unit, i e, 9 books can be arranged in 9! ways. Also 4 blue books can be arranged in 4! ways.

total number of combinations which have have all four books together= 9! 4!

since these 4 books are of same colour, so total number of combinations which have have all four blue books together$\displaystyle = \frac{9! \times 4!}{4!} = 9!$

7. Originally Posted by Shyam
so total number of combinations which have have all four blue books together$\displaystyle = \frac{9! \times 4!}{4!} = 9!$
This is a quite nice reasoning, but why are you dividing by 4! at the end? This amounts to considering the blue books as unordered, which is not the case. (or am I missing something ?)