Page 1 of 2 12 LastLast
Results 1 to 25 of 26

Thread: permutation help!

  1. #1
    Junior Member
    Join Date
    Jul 2012
    HSC
    2012
    Gender
    Male
    Posts
    179
    Rep Power
    5

    permutation help!

    can someone explain and show me the working out of this question?

    If the letters of the word GUMTREE and the letters of the word KOALA are combined
    and arranged into a single twelve-letter word, in how many of these arrangements do the
    letters of KOALA appear in their correct order, but not necessarily together?

    thanks!

    and will it help if i keep doing extension questions from cambridge textbook? they are extremely hard!!
    Last edited by john-doe; 30 Jul 2012 at 7:34 PM.

  2. #2
    This too shall pass Sy123's Avatar
    Join Date
    Nov 2011
    HSC
    2013
    Gender
    Male
    Posts
    3,734
    Rep Power
    8

    Re: permutation help!



    Simply treat the word KOALA as a whole, then find the number of different arrangements of this 'seven' element word. The reason you dont have a multiplier of 5! is that we dont want KOALA to have a different order. So we treat it so it cant change order, hence our answer.

  3. #3
    what is that?It is Cowpea RealiseNothing's Avatar
    Join Date
    Jul 2011
    HSC
    2013
    Gender
    Male
    Location
    Sydney
    Posts
    4,609
    Rep Power
    9

    Re: permutation help!

    Quote Originally Posted by Sy123 View Post


    Simply treat the word KOALA as a whole, then find the number of different arrangements of this 'seven' element word. The reason you dont have a multiplier of 5! is that we dont want KOALA to have a different order. So we treat it so it cant change order, hence our answer.
    You can't just treat the word KOALA as a whole.

  4. #4
    This too shall pass Sy123's Avatar
    Join Date
    Nov 2011
    HSC
    2013
    Gender
    Male
    Posts
    3,734
    Rep Power
    8

    Re: permutation help!

    Oh hang on I didnt see the other part saying not necessarily together. My bad.

  5. #5
    what is that?It is Cowpea RealiseNothing's Avatar
    Join Date
    Jul 2011
    HSC
    2013
    Gender
    Male
    Location
    Sydney
    Posts
    4,609
    Rep Power
    9

    Re: permutation help!

    I got 4324320.

    The working out is going to be very hard to explain and very tedious to type out, so I want to make sure that it's correct before actually writing out the solution. So is this right?

  6. #6
    Executive Member bleakarcher's Avatar
    Join Date
    Jul 2011
    HSC
    2013
    Gender
    Male
    Posts
    1,511
    Rep Power
    7

    Re: permutation help!

    I remember attempting this question lol, I was going to explode.
    Physics is to mathematics like sex is to masturbation.” —Richard Feynman

  7. #7
    what is that?It is Cowpea RealiseNothing's Avatar
    Join Date
    Jul 2011
    HSC
    2013
    Gender
    Male
    Location
    Sydney
    Posts
    4,609
    Rep Power
    9

    Re: permutation help!

    Quote Originally Posted by bleakarcher View Post
    I remember attempting this question lol, I was going to explode.
    The way to do this question is using the idea of triangular numbers like I told you about a while back. But for this question you have to extend the triangular numbers to summations of triangular numbers than the summation of the summation of triangular numbers etc lol.

  8. #8
    Executive Member bleakarcher's Avatar
    Join Date
    Jul 2011
    HSC
    2013
    Gender
    Male
    Posts
    1,511
    Rep Power
    7

    Re: permutation help!

    Quote Originally Posted by RealiseNothing View Post
    I got 4324320.

    The working out is going to be very hard to explain and very tedious to type out, so I want to make sure that it's correct before actually writing out the solution. So is this right?
    The answer is 1 995 840 according to my textbook.
    Physics is to mathematics like sex is to masturbation.” —Richard Feynman

  9. #9
    Executive Member bleakarcher's Avatar
    Join Date
    Jul 2011
    HSC
    2013
    Gender
    Male
    Posts
    1,511
    Rep Power
    7

    Re: permutation help!

    Quote Originally Posted by RealiseNothing View Post
    The way to do this question is using the idea of triangular numbers like I told you about a while back. But for this question you have to extend the triangular numbers to summations of triangular numbers than the summation of the summation of triangular numbers etc lol.
    YES FUCK!! it gets confusing as shit lol. thanks for teaching me that man. it has actually come in handy.
    Physics is to mathematics like sex is to masturbation.” —Richard Feynman

  10. #10
    what is that?It is Cowpea RealiseNothing's Avatar
    Join Date
    Jul 2011
    HSC
    2013
    Gender
    Male
    Location
    Sydney
    Posts
    4,609
    Rep Power
    9

    Re: permutation help!

    Quote Originally Posted by bleakarcher View Post
    The answer is 1 995 840 according to my textbook.
    lol ill go check my working out, I'll probably made a silly calculation effor somewhere since I'm 99% sure my method is right.

  11. #11
    Executive Member bleakarcher's Avatar
    Join Date
    Jul 2011
    HSC
    2013
    Gender
    Male
    Posts
    1,511
    Rep Power
    7

    Re: permutation help!

    Quote Originally Posted by RealiseNothing View Post
    lol ill go check my working out, I'll probably made a silly calculation effor somewhere since I'm 99% sure my method is right.
    cool, im gunna begin a second (hopefully successful) attempt at this problem.
    Physics is to mathematics like sex is to masturbation.” —Richard Feynman

  12. #12
    Junior Member
    Join Date
    Jul 2012
    HSC
    2012
    Gender
    Male
    Posts
    179
    Rep Power
    5

    Re: permutation help!

    But do u guys think that they will ask these difficult questions in the hsc?

  13. #13
    what is that?It is Cowpea RealiseNothing's Avatar
    Join Date
    Jul 2011
    HSC
    2013
    Gender
    Male
    Location
    Sydney
    Posts
    4,609
    Rep Power
    9

    Re: permutation help!

    Quote Originally Posted by bleakarcher View Post
    YES FUCK!! it gets confusing as shit lol. thanks for teaching me that man. it has actually come in handy.
    Basically this is what I've found, and I'm pretty sure it works:

    For questions where you want to find how many ways letters can be arranged so that the letters are in a certain order (ie the letter "a" precedes the letter "b"), you can use the following:

    Consider you wanted to arranged 'n' letters and you want 'k' of those letters to be in a particular order, like the question in this thread (the letters of koala need to be in order, so in this case n=12 and k=5). Then the following is what you do (I think):

    For k=2, amount of arrangements is:

    which is just the '(n-1)'th triangular number.

    This can also be written as:



    For k=3, amount of arrangements is:



    For k=4, amount of arrangements is:



    For k=5, amount of arrangements is:



    Etc.

    I'm going to go through this post and make sure I haven't did a typo or something and make sure everything is correct as well.

    But notice how the number on top of the summations always goes down by 1? It goes from n-1, to n-2, to n-3, to n-4, etc. And you just keep adding a summation each time basically.
    Last edited by RealiseNothing; 30 Jul 2012 at 9:25 PM.

  14. #14
    what is that?It is Cowpea RealiseNothing's Avatar
    Join Date
    Jul 2011
    HSC
    2013
    Gender
    Male
    Location
    Sydney
    Posts
    4,609
    Rep Power
    9

    Re: permutation help!

    Quote Originally Posted by john-doe View Post
    But do u guys think that they will ask these difficult questions in the hsc?
    No.

  15. #15
    This too shall pass Sy123's Avatar
    Join Date
    Nov 2011
    HSC
    2013
    Gender
    Male
    Posts
    3,734
    Rep Power
    8

    Re: permutation help!

    Quote Originally Posted by RealiseNothing View Post
    Basically this is what I've found, and I'm pretty sure it works:

    For questions where you want to find how many ways letters can be arranged so that the letters are in a certain order (ie the letter "a" precedes the letter "b"), you can use the following:

    Consider you wanted to arranged 'n' letters and you want 'k' of those letters to be in a particular order, like the question in this thread (the letters of koala need to be in order, so in this case n=12 and k=5). Then the following is what you do (I think):

    For k=2, amount of arrangements is:

    which is just the '(n-1)'th triangular number.

    This can also be written as:



    For k=3, amount of arrangements is:



    For k=4, amount of arrangements is:



    For k=5, amount of arrangements is:



    Etc.

    I'm going to go through this post and make sure I haven't did a typo or something and make sure everything is correct as well.

    But notice how the number on top of the summations always goes down by 1? It goes from n-1, to n-2, to n-3, to n-4, etc. And you just keep adding a summation each time basically.
    This method is ingenious...

    Can you explain why this works though?

  16. #16
    Executive Member bleakarcher's Avatar
    Join Date
    Jul 2011
    HSC
    2013
    Gender
    Male
    Posts
    1,511
    Rep Power
    7

    Re: permutation help!

    Quote Originally Posted by RealiseNothing View Post
    Basically this is what I've found, and I'm pretty sure it works:

    For questions where you want to find how many ways letters can be arranged so that the letters are in a certain order (ie the letter "a" precedes the letter "b"), you can use the following:

    Consider you wanted to arranged 'n' letters and you want 'k' of those letters to be in a particular order, like the question in this thread (the letters of koala need to be in order, so in this case n=12 and k=5). Then the following is what you do (I think):

    For k=2, amount of arrangements is:

    which is just the '(n-1)'th triangular number.

    This can also be written as:



    For k=3, amount of arrangements is:



    For k=4, amount of arrangements is:



    For k=5, amount of arrangements is:



    Etc.

    I'm going to go through this post and make sure I haven't did a typo or something and make sure everything is correct as well.

    But notice how the number on top of the summations always goes down by 1? It goes from n-1, to n-2, to n-3, to n-4, etc. And you just keep adding a summation each time basically.
    how awesome..
    Physics is to mathematics like sex is to masturbation.” —Richard Feynman

  17. #17
    what is that?It is Cowpea RealiseNothing's Avatar
    Join Date
    Jul 2011
    HSC
    2013
    Gender
    Male
    Location
    Sydney
    Posts
    4,609
    Rep Power
    9

    Re: permutation help!

    Quote Originally Posted by Sy123 View Post
    This method is ingenious...

    Can you explain why this works though?
    I'll use an example, it'll be much easier to see I think with an example.

    Take the letters of the word MATHS and arrange them so the 'A' always comes after the 'M'.

    Fix the 'M' in the first spot:

    M A _ _ _

    M _ A _ _

    M _ _ A _

    M _ _ _ A

    So there are 4 ways to do this, now fix the M in the 2nd spot:

    _ M A _ _

    _ M _ A _

    _ M _ _ A

    So there are 3 ways to do this, now fix the 'M' in the 3rd spot:

    _ _ M A _

    _ _ M _ A

    So there are 2 ways, and finally fix the 'M' in the 4th spot:

    _ _ _ M A

    So there is only 1 way.

    Amount of ways = 1 + 2 + 3 + 4 = 10

    This is just the 4th triangular number. If we used the letters ABCDEF and had B after A, it would be the 5th triangular number, and so on, hence it is the (n-1)th triangular number when there are n letters.

    Now consider MATHS again, and say we want A to be after M, and T to be after A, so in this order - MAT:

    Fix M in the first spot:

    M A T _ _

    M A _ T _

    M A _ _ T

    M _ A T _

    M _ A _ T

    M _ _ A T

    Which is 6 ways.

    So you should be able to see that when we fix M in the first spot, there are now only 4 letters instead of 5, and we want to arrange 2 letters again (A and T) like befor, so we can use the (n-1) triangular number, so since n is now 4, we have the 3rd triangular number, which is 1+2+3=6 as required.

    If we fix M in the 2nd spot we get:

    _ M _ _ _

    So if we put A and T in there, the amount of ways will be the (n-1) triangular number, and we know there are only 3 spaces to put the A and T, so 'n' is now 3, hence it is the 2nd triangular number, that is 1+2=3.

    If we fix M in the 3rd position we get:

    _ _ M _ _

    And now there are only 2 letters, so using (n-1) it is the 1st triangular number, hence there is only 1 arrangement with M in the 3rd spot (which should be obvious, it is _ _ M A T).

    So the total arrangements when A comes before M, and T comes before A (M A T), is 6 + 3 + 1 = 3rd triangular number + 2nd triangular number + 1st triangular number = Sum of the first (n-2) triangular numbers (remember that 'n' is the total amount of letters in the word, in this case MATHS, so n=5, (n-2)=3, sum of first 3 triangular numbers).

    If I then wanted it so that 4 letters are in order, for example M A T H in that oder, it would just become the "sum of the sum of the (n-3) triangular numbers) since it will follow the same pattern as before.

    Hope this makes sense lol.

  18. #18
    This too shall pass Sy123's Avatar
    Join Date
    Nov 2011
    HSC
    2013
    Gender
    Male
    Posts
    3,734
    Rep Power
    8

    Re: permutation help!

    Quote Originally Posted by RealiseNothing View Post
    I'll use an example, it'll be much easier to see I think with an example.

    Take the letters of the word MATHS and arrange them so the 'A' always comes after the 'M'.

    Fix the 'M' in the first spot:

    M A _ _ _

    M _ A _ _

    M _ _ A _

    M _ _ _ A

    So there are 4 ways to do this, now fix the M in the 2nd spot:

    _ M A _ _

    _ M _ A _

    _ M _ _ A

    So there are 3 ways to do this, now fix the 'M' in the 3rd spot:

    _ _ M A _

    _ _ M _ A

    So there are 2 ways, and finally fix the 'M' in the 4th spot:

    _ _ _ M A

    So there is only 1 way.

    Amount of ways = 1 + 2 + 3 + 4 = 10

    This is just the 4th triangular number. If we used the letters ABCDEF and had B after A, it would be the 5th triangular number, and so on, hence it is the (n-1)th triangular number when there are n letters.

    Now consider MATHS again, and say we want A to be after M, and T to be after A, so in this order - MAT:

    Fix M in the first spot:

    M A T _ _

    M A _ T _

    M A _ _ T

    M _ A T _

    M _ A _ T

    M _ _ A T

    Which is 6 ways.

    So you should be able to see that when we fix M in the first spot, there are now only 4 letters instead of 5, and we want to arrange 2 letters again (A and T) like befor, so we can use the (n-1) triangular number, so since n is now 4, we have the 3rd triangular number, which is 1+2+3=6 as required.

    If we fix M in the 2nd spot we get:

    _ M _ _ _

    So if we put A and T in there, the amount of ways will be the (n-1) triangular number, and we know there are only 3 spaces to put the A and T, so 'n' is now 3, hence it is the 2nd triangular number, that is 1+2=3.

    If we fix M in the 3rd position we get:

    _ _ M _ _

    And now there are only 2 letters, so using (n-1) it is the 1st triangular number, hence there is only 1 arrangement with M in the 3rd spot (which should be obvious, it is _ _ M A T).

    So the total arrangements when A comes before M, and T comes before A (M A T), is 6 + 3 + 1 = 3rd triangular number + 2nd triangular number + 1st triangular number = Sum of the first (n-2) triangular numbers (remember that 'n' is the total amount of letters in the word, in this case MATHS, so n=5, (n-2)=3, sum of first 3 triangular numbers).

    If I then wanted it so that 4 letters are in order, for example M A T H in that oder, it would just become the "sum of the sum of the (n-3) triangular numbers) since it will follow the same pattern as before.

    Hope this makes sense lol.
    Thanks for taking the time to post it up heh, I get the concept now.

  19. #19
    Member
    Join Date
    Jul 2010
    HSC
    2009
    Gender
    Male
    Posts
    423
    Rep Power
    7

    Re: permutation help!

    1) Since you want the letters of KOALA in order, fix those first:
    __ K __ O __ A __ L __ A __
    The lines indicate the spots the letters of GUMTREE can be placed

    2) Choose the spot for the letter G: 6 spots means 6 ways to do so. Then our arrangement becomes something like (depending on your spot for G):
    __ K __ O __ A __ L __ G __ A __

    3) Choose the spot for the 2nd letter: Now there are 7 spots and hence 7 ways to do so. Then our arrangement becomes something like:
    __ K __U __ O __ A __ L __ G __ A __

    Continue this above pattern for each letter of GUMTREE, and then dividing by 2 since the letter E is repeated, we get the number of ways to be: 6*7*8*9*10*11*12/2=1995840

  20. #20
    Cadet
    Join Date
    Jul 2012
    HSC
    2011
    Gender
    Male
    Posts
    42
    Rep Power
    6

    Re: permutation help!

    No. of arrangements= (7!/2!) x [ 6 + 6C2 (2!+2!+2!) + 6C3 ( 3!/2! + 3! + 3!/2! + 3!/2!) + 6C4 (4!/3!+4!/2!+4!/3!) + 6C5 ( 5!/4! + 5!/3!2! ) + 6C6 x 6!/5!]

    dumb question because its unnecessarily long

    EDIT: Almost forgot last bracket

    EDIT #2: that long thing evaluates to 1995840
    Last edited by DAFUQ; 31 Jul 2012 at 2:29 AM.

  21. #21
    Cadet
    Join Date
    Jul 2012
    HSC
    2011
    Gender
    Male
    Posts
    42
    Rep Power
    6

    Re: permutation help!

    Quote Originally Posted by deterministic View Post
    1) Since you want the letters of KOALA in order, fix those first:
    __ K __ O __ A __ L __ A __
    The lines indicate the spots the letters of GUMTREE can be placed

    2) Choose the spot for the letter G: 6 spots means 6 ways to do so. Then our arrangement becomes something like (depending on your spot for G):
    __ K __ O __ A __ L __ G __ A __

    3) Choose the spot for the 2nd letter: Now there are 7 spots and hence 7 ways to do so. Then our arrangement becomes something like:
    __ K __U __ O __ A __ L __ G __ A __

    Continue this above pattern for each letter of GUMTREE, and then dividing by 2 since the letter E is repeated, we get the number of ways to be: 6*7*8*9*10*11*12/2=1995840
    i think thats the simplest method ive seen so far. +1 like

  22. #22
    Ancient Orator 4025808's Avatar
    Join Date
    Apr 2009
    HSC
    2011
    Uni Grad
    2017
    Gender
    Male
    Location
    中國農村稻農
    Posts
    4,380
    Rep Power
    10

    Re: permutation help!

    Quote Originally Posted by deterministic View Post
    1) Since you want the letters of KOALA in order, fix those first:
    __ K __ O __ A __ L __ A __
    The lines indicate the spots the letters of GUMTREE can be placed

    2) Choose the spot for the letter G: 6 spots means 6 ways to do so. Then our arrangement becomes something like (depending on your spot for G):
    __ K __ O __ A __ L __ G __ A __

    3) Choose the spot for the 2nd letter: Now there are 7 spots and hence 7 ways to do so. Then our arrangement becomes something like:
    __ K __U __ O __ A __ L __ G __ A __

    Continue this above pattern for each letter of GUMTREE, and then dividing by 2 since the letter E is repeated, we get the number of ways to be: 6*7*8*9*10*11*12/2=1995840
    Probably the best sort of thinking to approach this Q. Just make sure your answer is right too. +1
    B Engineering (Petroleum) / B Science (Mathematics, Statistics) @ UNSW

    For Maths Tutoring inquiries, please visit here: http://community.boredofstudies.org/...d.php?t=281590

  23. #23
    Cadet
    Join Date
    Jul 2012
    HSC
    2011
    Gender
    Male
    Posts
    42
    Rep Power
    6

    Re: permutation help!

    Quote Originally Posted by 4025808 View Post
    Probably the best sort of thinking to approach this Q. Just make sure your answer is right too. +1
    it should be right because i also did it with a different method

  24. #24
    Ancient Orator 4025808's Avatar
    Join Date
    Apr 2009
    HSC
    2011
    Uni Grad
    2017
    Gender
    Male
    Location
    中國農村稻農
    Posts
    4,380
    Rep Power
    10

    Re: permutation help!

    Quote Originally Posted by DAFUQ View Post
    it should be right because i also did it with a different method
    Oh okay, well we'll see what the answer says. OP, what's the answer?
    B Engineering (Petroleum) / B Science (Mathematics, Statistics) @ UNSW

    For Maths Tutoring inquiries, please visit here: http://community.boredofstudies.org/...d.php?t=281590

  25. #25
    Junior Member
    Join Date
    Jul 2012
    HSC
    2012
    Gender
    Male
    Posts
    179
    Rep Power
    5

    Re: permutation help!

    And the answer is 1995840 !!!!! Thanks everyone!!!!!!!

Page 1 of 2 12 LastLast

Thread Information

Users Browsing this Thread

There are currently 1 users browsing this thread. (0 members and 1 guests)

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •