Permutations with restrictions

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[Voon Nyuk Fung]

9 different books are to be arranged on a book shelf. 6 of these books were written by Shakespeare and 3 by Dickens. How many possible permutations are there if
(a) the books by Dickens must be next to each other?
(b) the books by Dickens are separated from each other?

Can anyone help to solve part (b) so that a student would be able to understand it?
Thank you.

[Voon Nyuk Fung]

(a) To ensure that books by Dickens are together, we gather them together to form one block. Within this block, the three books by Dickens can be permuted in 3! ways.
This block with the remaining 6 books form 7 objects which can be permuted in 7! ways.
Hence by the basic counting principle, the number of permutations = 3! X 7!

There seems to be no such simple argument for part (b). Please help, thank you.

[Lim Yu Teong]

The answer for (b) is 6! x 7P3. Hope that is what you want. Have a nice and wonderful week.

[Wong Tok Hua]

There are 6! ways to arrange Shakespears books. And all Dicken's books must be seperated. Which means that those 3 Dicken's books must be placed in between Shakespear's books or at the two ends of the row of books. That means we have 7 spaces for those 3 Dicken's books. So the number of arrangements for Dicken's books is 7P3. And therefore the answer is 6! x 7P3.

Hope this explanation helps.

[Zefry Hanif Borhan]

so which one the right answer for (b)?

[Tan Pack Lang]

Thanks very much, Mr. Lim and Mr. Wong. I have to admit i have the wrong concept. I'll delete off my posting to avoid confusion.

[Goh Yong Juan]

so which one the right answer for (b)?

6! x 7P3 is the right answer.

Suggestion : complicated situation can be visualised by using diagrams.

[Tan Pack Lang]

That's very good. Maybe Cikgu Goh can show us the visualisation or diagram by uploading the pictures of your visualisation. Thanks so much....

[Voon Nyuk Fung]

P1 P2 P3 P4 P5 P6 P7 P8 P9 (P-position)
C1 S  C2 S  C3 S   S  S  S   Number of arrangements = 6! x 3! x 5 (C3 can take any of the 5 positions from P5 to P9)
C1 S  S  C2 S  C3  S  S  S   Number of arrangements = 6! x 3! x 4

By considering other cases, the total number of arrangements = 6! x 3! x35
This is long and time consuming.
Thank you to Mr Lim and Mr Wong for your help. Appreciate it very much.
Have a happy holiday.

[Goh Yong Juan]

That's very good. Maybe Cikgu Goh can show us the visualisation or diagram by uploading the pictures of your visualisation. Thanks so much....

Hi Mr. Tan I attached the diagram. Please comment. Good day. Thank you.

Sorry I cant attached the diagram (cant upload the word file) the following is the simplify one:


 D  S  D  S  D  S  D  S  D  S  D  S  D


From the diagram, if Dicken (D) must be separated there are 7 possible places for Dicken (arrow). Since the order is important then Dicken will have 7 P 3 ways to be arranged.

Shakespeare (S) has 6! Ways.

Total = 6!  X 7 P 3

OK

Thank you.

[Tan Pack Lang]

Thanks very much, Mr. Goh.

[Mohd Rizuan Mohd Nor]

By using excel, from 7P3 = 210 ways, there are 6 set of 35 different arrangement. Verify that 6! x 7P3 is the correct answer. I need to workout to understand this as fast as possible since that by using diagram consuming  a lot of time.

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