Thread: 10 books are arranged on a shelf randomly...

1. 10 books are arranged on a shelf randomly...

10 books are arranged on a shelf randomly.

A) What's the probability that between the books "Cinderella" and "The Hobbit" 3 books exactly will seperate them?

I tried to do 9!*4!*2! divided by 10!

But..the answer doesn't make sense, that's 480%!

Help?

B) Among the 10 books, 4 have red covers, 4 with blue and 2 with white. What's the probability that all books from the same colour are next to each other?

Here I tried to do 3!*4!*4*2! = 6912

Then 3! = 6 (because the order of what colours doesn't matter)

then 6912*6 = 41472

Then 41472 divided by 10! (which is all possible arrangements) I get = 0.011

Makes sense?

-nerd_girl (AKA April)

2. A) C is Cinderella and H is Hobbit. The subset CBBBH has 5 elements and there are 6 possible locations for them in the set of 10 books, and 8! ways to arrange the remaining 8 books for each of these 6 positions

CBBBHBBBBB
BCBBBHBBBB
BBCBBBHBBB
BBBCBBBHBB
BBBBCBBBHB
BBBBBCBBBH

so that there are 8!*6*2 = 483840 ways 3 books can separate C and H (C and H can be interchanged, hence factor of 2). Then

P(3 books in between C and H) = 483840 / 10! = .134

B) I think you included a factor of 6 twice on accident. Divide your answer by 6 and it should be good.

3. Should we also consider HBBBC?
Originally Posted by zmogggggg
A) C is Cinderella and H is Hobbit. The subset CBBBH has 5 elements and there are 6 possible locations for them in the set of 10 books, and 8! ways to arrange the remaining 8 books for each of these 6 positions

CBBBHBBBBB
BCBBBHBBBB
BBCBBBHBBB
BBBCBBBHBB
BBBBCBBBHB
BBBBBCBBBH

so that there are 8!*6*2 = 483840 ways 3 books can separate C and H (C and H can be interchanged, hence factor of 2). Then

P(3 books in between C and H) = 483840 / 10! = .134

B) I think you included a factor of 6 twice on accident. Divide your answer by 6 and it should be good.

4. Originally Posted by semic
Should we also consider HBBBC?
The factor of 2 considers this.

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