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Number of permutations such that each prime is followed by at least one composite

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1

$begingroup$

Question: Let $P$ be the number of permutations of $3,4,5,6,7,8$ such that each prime is followed by at least one composite number. Find $P$.

Using the string method I got $36$ as the answer and by another approach I got $1440$ as the answer. I do not think that these are the correct answers. Please help.

combinatorics permutations

share|cite|improve this question

edited Mar 13 at 18:30

6005

37k751127

asked Mar 13 at 18:04

MrAPMrAP

1,17421432

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The primes are 3, 5 and 7. Note that the condition that a prime is followed by a composite number is equivalent to the primes occupying the first, third and fifth positions, while the three composite numbers occupy the second, fourth and sixth positions (Else we wold get adjacent pairs of primes). Thus we can realise all such permutations by permuting the three primes among themselves and permuting the three composite numbers among themselves, and we get $3!*3!=36$ possible permutations.
  $endgroup$
  – Heinrich Wagner
  Mar 13 at 18:14
  

  
  
  
  

add a comment | 

1

$begingroup$

Question: Let $P$ be the number of permutations of $3,4,5,6,7,8$ such that each prime is followed by at least one composite number. Find $P$.

Using the string method I got $36$ as the answer and by another approach I got $1440$ as the answer. I do not think that these are the correct answers. Please help.

combinatorics permutations

share|cite|improve this question

edited Mar 13 at 18:30

6005

37k751127

asked Mar 13 at 18:04

MrAPMrAP

1,17421432

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The primes are 3, 5 and 7. Note that the condition that a prime is followed by a composite number is equivalent to the primes occupying the first, third and fifth positions, while the three composite numbers occupy the second, fourth and sixth positions (Else we wold get adjacent pairs of primes). Thus we can realise all such permutations by permuting the three primes among themselves and permuting the three composite numbers among themselves, and we get $3!*3!=36$ possible permutations.
  $endgroup$
  – Heinrich Wagner
  Mar 13 at 18:14
  

  
  
  
  

add a comment | 

$begingroup$

Question: Let $P$ be the number of permutations of $3,4,5,6,7,8$ such that each prime is followed by at least one composite number. Find $P$.

Using the string method I got $36$ as the answer and by another approach I got $1440$ as the answer. I do not think that these are the correct answers. Please help.

combinatorics permutations

share|cite|improve this question

edited Mar 13 at 18:30

6005

37k751127

asked Mar 13 at 18:04

MrAPMrAP

1,17421432

$endgroup$

share|cite|improve this question

edited Mar 13 at 18:30

6005

37k751127

asked Mar 13 at 18:04

MrAPMrAP

1,17421432

share|cite|improve this question

edited Mar 13 at 18:30

6005

37k751127

asked Mar 13 at 18:04

MrAPMrAP

1,17421432

edited Mar 13 at 18:30

6005

37k751127

edited Mar 13 at 18:30

6005

37k751127

asked Mar 13 at 18:04

MrAPMrAP

1,17421432

asked Mar 13 at 18:04

MrAPMrAP

1,17421432

1 Answer
1

active

oldest

votes

1

$begingroup$

To tackle this problem, we can start with the condition:

each prime is followed by at least one composite.

I think Vinyl_cape_jawa has interpreted this incorrectly -- it means that each prime is followed by at least one composite somewhere on the list. But this is equivalent to saying that the last prime is followed by at least one composite. And all of these numbers are either composite or prime.

So it's actually just saying that: the list ends in a composite!

So: how many permutations are there such that the list ends in a composite?
Start by finding the number of ways to choose the last element; then find the number of ways to order all the remaining elements.

P.S.
I haven't heard of the "string method", but I think that neither of your two answers are right.

$1440$ can't be right because the total number of permutations of the set is $6! = 720$, which is smalle.r

$36$ I think would be correct if Vinyl_cape_jawa's interpretation is right.

share|cite|improve this answer

edited Mar 15 at 14:59

answered Mar 13 at 18:33

60056005

37k751127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @N.F.Taussig Yeah I mixed that up. Fixed now.
  $endgroup$
  – 6005
  Mar 15 at 14:59
  

  
  
  
  

add a comment | 

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1

$begingroup$

To tackle this problem, we can start with the condition:

each prime is followed by at least one composite.

I think Vinyl_cape_jawa has interpreted this incorrectly -- it means that each prime is followed by at least one composite somewhere on the list. But this is equivalent to saying that the last prime is followed by at least one composite. And all of these numbers are either composite or prime.

So it's actually just saying that: the list ends in a composite!

So: how many permutations are there such that the list ends in a composite?
Start by finding the number of ways to choose the last element; then find the number of ways to order all the remaining elements.

P.S.
I haven't heard of the "string method", but I think that neither of your two answers are right.

$1440$ can't be right because the total number of permutations of the set is $6! = 720$, which is smalle.r

$36$ I think would be correct if Vinyl_cape_jawa's interpretation is right.

share|cite|improve this answer

edited Mar 15 at 14:59

answered Mar 13 at 18:33

60056005

37k751127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @N.F.Taussig Yeah I mixed that up. Fixed now.
  $endgroup$
  – 6005
  Mar 15 at 14:59
  

  
  
  
  

add a comment | 

1

$begingroup$

To tackle this problem, we can start with the condition:

each prime is followed by at least one composite.

I think Vinyl_cape_jawa has interpreted this incorrectly -- it means that each prime is followed by at least one composite somewhere on the list. But this is equivalent to saying that the last prime is followed by at least one composite. And all of these numbers are either composite or prime.

So it's actually just saying that: the list ends in a composite!

So: how many permutations are there such that the list ends in a composite?
Start by finding the number of ways to choose the last element; then find the number of ways to order all the remaining elements.

P.S.
I haven't heard of the "string method", but I think that neither of your two answers are right.

$1440$ can't be right because the total number of permutations of the set is $6! = 720$, which is smalle.r

$36$ I think would be correct if Vinyl_cape_jawa's interpretation is right.

share|cite|improve this answer

edited Mar 15 at 14:59

answered Mar 13 at 18:33

60056005

37k751127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @N.F.Taussig Yeah I mixed that up. Fixed now.
  $endgroup$
  – 6005
  Mar 15 at 14:59
  

  
  
  
  

add a comment | 

$begingroup$

To tackle this problem, we can start with the condition:

each prime is followed by at least one composite.

I think Vinyl_cape_jawa has interpreted this incorrectly -- it means that each prime is followed by at least one composite somewhere on the list. But this is equivalent to saying that the last prime is followed by at least one composite. And all of these numbers are either composite or prime.

So it's actually just saying that: the list ends in a composite!

So: how many permutations are there such that the list ends in a composite?
Start by finding the number of ways to choose the last element; then find the number of ways to order all the remaining elements.

P.S.
I haven't heard of the "string method", but I think that neither of your two answers are right.

$1440$ can't be right because the total number of permutations of the set is $6! = 720$, which is smalle.r

$36$ I think would be correct if Vinyl_cape_jawa's interpretation is right.

share|cite|improve this answer

edited Mar 15 at 14:59

answered Mar 13 at 18:33

60056005

37k751127

$endgroup$

share|cite|improve this answer

edited Mar 15 at 14:59

answered Mar 13 at 18:33

60056005

37k751127

answered Mar 13 at 18:33

60056005

37k751127

answered Mar 13 at 18:33

60056005

37k751127