# arrange book's question

• Jun 28th 2007, 11:47 PM
neworld222
arrange book's question
places on at will the bookshelf 10 books, asks 5 books which in which assigns to put on the same place probability
• Jun 29th 2007, 05:48 AM
galactus
What is that supposed to mean?. I am sorry, but it makes no sense.
• Jun 29th 2007, 08:04 AM
Soroban
Hello, neworld222!

I think it's Yoda, the Jedi Master . . .

Quote:

places on at will the bookshelf 10 books,
asks 5 books which in which assigns to put on the same place probability

I'll take a wild guess . . .

Ten distinct books are placed randomly on a shelf (in a row).
What is the probability that a particular five books are adjacent?

There are $10!$ possible arrangements.

Duct-tape those five books together.
Then we have six "books" to arrange: . $\left\{\boxed{ABCDE},\,F,\,G,\,H,\,I,\,J\right\}$
. . They can be arranged in $6!$ ways.

For each of these, the five books can be permuted in $5!$ ways.
. . Hence, there are: . $(6!)(5!)$ desirable arrangements.

The probability is: . $\frac{(6!)(5!)}{10!} \;=\;\frac{1}{42}$

• Jun 30th 2007, 01:49 AM
neworld222
Quote:

Originally Posted by Soroban
Hello, neworld222!

I think it's Yoda, the Jedi Master . . .

I'll take a wild guess . . .

Ten distinct books are placed randomly on a shelf (in a row).
What is the probability that a particular five books are adjacent?

There are $10!$ possible arrangements.

Duct-tape those five books together.
Then we have six "books" to arrange: . $\left\{\boxed{ABCDE},\,F,\,G,\,H,\,I,\,J\right\}$
. . They can be arranged in $6!$ ways.

For each of these, the five books can be permuted in $5!$ ways.
. . Hence, there are: . $(6!)(5!)$ desirable arrangements.

The probability is: . $\frac{(6!)(5!)}{10!} \;=\;\frac{1}{42}$

yes ,you give what i need .English is not my first lanuage,so i sorry for that i made you take a wild guess