Permutation And Combination

Probability Level 2

9 different books are to be arranged on a bookshelf. 4 of these books were written by Shakespeare, 2 by Dickens, and 3 by Conrad. How many possible permutations are there if the books by Conrad must be separated from one another?

23331 67800 151200 44443310

2 solutions

J J
Jun 5, 2016

The number of ways to order Shakespeare's and Dickens' book is $6!$

There will be 7 spaces (in front of all the books, in between the first and second book, in between the second and third book...all the way until at the end of all the books) that Conrad's books can go with them since they can't be next to each other.

The number of ways of doing this is $7 \choose 3$ $\cdot 3!$ since once they are in their spots they can be ordered $3!$ ways.

Thus, the number of ways is $6! \cdot$ $7 \choose 3$ $\cdot 3!$ = $\boxed{151200}$ .

can't we firstly find out the total number arrangements of all the books and then subtract from it the number of ways in which conrad's books appear together by considering them one entity ????? (((9!= A= number of ways of arranging all the books , 7!*3!= B = number of ways in which conrad's books are together in which 3! is the arrangement of three books of conrad. now to get whatever we want ...A-B. IS IT WRONG???

Dolsi Varshney - 2 years, 4 months ago

It is also possible that only two of the three Conrad's books are next to each other, and you must also subtract these cases from the total number of arrangements to find the number of ways Conrad's books are all separate.

J J - 2 years, 3 months ago

please do tell

Dolsi Varshney - 2 years, 4 months ago

It's a valid solution but it's incomplete, you are not considering the cases where two books are adjacent but the third is away, i.e. CCSSSSDDC is considered a valid combination in your reasoning, when it is not in reality.

Sonoda Taiga - 2 years ago

Dolsi: in your solution you forgot to consider the scenario when only two Conrad books were together.

Tim Tso - 1 year, 10 months ago

the JJ's logic seems to be correct, but doing the calculus something is wrong: 6! = 720 7! / (7-3)! = 210 3! = 6 720 * 210 * 6 = 907200 which is wrong... but if we exclude the 3! the result is correct (720 * 210 = 151200)

Dario Randazzo - 1 year, 7 months ago

Hmm it seems you mixed up 7 choose 3 with 7!/(7-3)!. 7 choose 3 is actually 7!/(3!(7-3)!)

J J - 1 year, 7 months ago

oh yeah, thank you!

Dario Randazzo - 1 year, 7 months ago

@Dario Randazzo – No problem!

J J - 1 year, 7 months ago

R.R Aa
Apr 21, 2020

Consider the case where Conrad's book are all adjacent, there are 7 such case and for each case 6! to place the other books, and 3! ways to arrange Conrad's books, so there are: 42*6! possibilities.

The second case is where only 2 of the 3 books are adjacent, there are two sub cases:

a) the two books are adjacent on the boundary of the bookshelf, then there are 2 of those cases, in each case 6! ways to place the other book and 7 ways to place the third book of Conrad thus we get: 72*6!
b) the two books are adjacent but none of them belongs to the boundary of the shelf, we get: 180*6! The result is then: 9!-294(6!)=151200

Permutation And Combination

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