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# Permutations and combination watch

1. There are eight different books on a bookshelf: three of them are hardbacks and the rest are paperbacks.

a) in how many different ways can the book be arranged if all the paperbacks are together and all the hardbacks are together?

b) in how many different ways can the books be arranged if all the paperbacks are together?
2. (Original post by magimagi1017)
There are eight different books on a bookshelf: three of them are hardbacks and the rest are paperbacks.

a) in how many different ways can the book be arranged if all the paperbacks are together and all the hardbacks are together?

b) in how many different ways can the books be arranged if all the paperbacks are together?
Hard to give a hint for this one.
The number of distinct permutations of n objects where there are p of one thing and q of another is:

Welcome to TSR btw
3. (Original post by joostan)
Hard to give a hint for this one.
The number of distinct permutations of n objects where there are p of one thing and q of another is:
Not sure how that helps.

(Original post by magimagi1017)
...
For a)

The paperbacks can be in any order, so how many is that?

Similarly the hardbacks.

Then either the paperbacks are before the hardbacks or the other way around - so multiply by 2.

b)

How many ways to arrange all the paperbacks?

Then treating the paperbacks as one item, how many ways to arrange all the items.
4. (Original post by ghostwalker)
Not sure how that helps.
5. it might help if you draw a picture of em,

there are three hardbacks, they can be arranged in 3! ways

there are 5 paperbacks, they can be arranged in 5! ways.

because they are clumped together, you treat the paperbacks and the hardbacks as one item, there are two items and they can be arranged in 2 ways.

hardbacks; paperbacks or paperback:hardbacks.

so you need to do 5!X3!X2

b) its a similar process, exept we dont need to separate the hardbacks, we can count them each as one item on the shelf. the paperbacks are grouped again so theyre 1, but they can be ordered in 5!

hardback; paperbacks; hardback:hardback or variations.

5!X4!
6. (Original post by VeeOMGlob)
it might help if you draw a picture of em,

there are three hardbacks, they can be arranged in 3! ways

there are 5 paperbacks, they can be arranged in 5! ways.

because they are clumped together, you treat the paperbacks and the hardbacks as one item, there are two items and they can be arranged in 2 ways.

hardbacks; paperbacks or paperback:hardbacks.

so you need to do 5!X3!X2

b) its a similar process, exept we dont need to separate the hardbacks, we can count them each as one item on the shelf. the paperbacks are grouped again so theyre 1, but they can be ordered in 5!

hardback; paperbacks; hardback:hardback or variations.

5!X4!
How do you get 4! in b), ?
7. (Original post by Aaqde)
How do you get 4! in b), ?
Since the person you're quoting was last active over a year and a half ago, you're unlikely to get a reply from them.

For b).

If all the paperbacks are together, the number of ways to arrange them is 5!

Now since they must be together treat them as one item amongst all the other books. So, there are 4 items, the block of paperbacks and the three hardbacks.
Number of arrangements of these 4 is 4!

So, total number of arrangements is 5!4!

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