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Math Help - permutations and factorials

  1. #1
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    permutations and factorials

    1. calculate the number of ways the letters of the word OLYMPICS can be arranged if:

    a) there are no restrictions
    b) the arrangment begins with L and ends with P
    c) the consonants are together (Y is a vowel here)
    d) the O and S are not together

    my thinking

    a) 8!
    b) 6!
    c)+d) i have no idea
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  2. #2
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    Quote Originally Posted by william View Post
    1. calculate the number of ways the letters of the word OLYMPICS can be arranged if:
    a) there are no restrictions
    b) the arrangment begins with L and ends with P
    c) the consonants are together (Y is a vowel here)
    d) the O and S are not together

    my thinking
    a) 8! CORRECT
    b) 6! CORRECT
    c)+d) i have no idea
    Three vowels and five consonants. Think of the consonants as one block that can be arranged in 5! ways.
    That block along with the three vowels give four blocks to arrange.
    How many ways do we get?

    For none of the vowels to be together they must be separated by the consonants.
    _L_M_P_C_S_ that is six places to put the vowels. {6 \choose 3}(3!)(5!)
    Now you explain those numbers.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by william View Post
    1. calculate the number of ways the letters of the word OLYMPICS can be arranged if:

    a) there are no restrictions
    b) the arrangment begins with L and ends with P
    c) the consonants are together (Y is a vowel here)
    d) the O and S are not together

    my thinking

    a) 8!
    Good!

    b) 6!
    Nice!

    c)
    i will give you a hint, see if you can come up with the answer.

    there are 5 consonants and 3 vowels. we are not told that the vowels have to be together, so they have more freedom and can actually be apart (i'm assuming)

    now, let c represent any of the consonants and v represent any of the vowels. there are 4 kinds of arrangements in which we can have the consonants together:

    cccccvvv
    vcccccvv
    vvcccccv
    vvvccccc

    d)
    Hint: how many arrangements can you find where they ARE together? find this, and then subtract it from your answer in (a) to get your solution (i hope you se why this works)
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  4. #4
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    Quote Originally Posted by Plato View Post
    Three vowels and five consonants. Think of the consonants as one block that can be arranged in 5! ways.
    That block along with the three vowels give four blocks to arrange.
    How many ways do we get?

    For none of the vowels to be together they must be separated by the consonants.
    _L_M_P_C_S_ that is six places to put the vowels. {6 \choose 3}(3!)(5!)
    Now you explain those numbers.
    Ok.. what is the notation {6 \choose 3} Sorry I never learned that. Secondly, I had an idea for c)

    let's say I let x=O,Y,I

    then i would have the 5 consonants and 4 vowels so couldn't I conclude then that it is 5!x4!=2880
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  5. #5
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    Hello, william;!

    1. Calculate the number of ways the letters of the word OLYMPICS can be arranged if:

    a) there are no restrictions
    b) the arrangment begins with L and ends with P
    c) the consonants are together (Y is a vowel here)
    d) the O and S are not together

    my thinking

    a) 8!
    b) 6!

    Right!
    c) The consonants are together.

    We have 5 consonants and 3 vowels.

    Duct-tape the consonants together.
    . . Then we have 4 "letters" to arrange: . \boxed{LMPCS},\,O,\,I,\,Y

    And there are 4! arrangements of the four "letters".

    But in each arrangement, the 5 consonants can be ordered in 5! ways.

    Therefore, there are: . 4! \times 5! \:=\:24 \times 120 \:=\:2,\!880 arrangements
    . . with the consonants together.



    d) O and S are not together.

    We know there are 8! possible arrangements.

    Let's count the arrangement in which O and S are together.


    Duct-tape the O and S together.
    . . Then we have seven "letters" to arrange: . \boxed{OS},\,L,\,Y,\,M,\,P,\,I,\,C

    And there are 7! arrangements of the seven "letters."

    But in each arrangement, the OS could be ordered SO.

    Hence, there are: . 2 \times 7! arrangements with O and S together.


    Therefore, there are: . 8! - 2\!\cdot\!7! \:=\:40,320 - 10,080 \:=\:30,240 arrangements
    . . with O and S nonadjacent.

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  6. #6
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    Quote Originally Posted by Jhevon View Post
    Hint: how many arrangements can you find where they ARE together? find this, and then subtract it from your answer in (a) to get your solution (i hope you se why this works)
    So for the total 8!=40320 and for O,S is 2!=2? I'm not quite getting it

    edit: this was created after seeing Sorobans post
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  7. #7
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    Quote Originally Posted by Soroban View Post
    Hello, william;!

    c) The consonants are together.

    We have 5 consonants and 3 vowels.

    Duct-tape the consonants together.
    . . Then we have 4 "letters" to arrange: . \boxed{LMPCS},\,O,\,I,\,Y

    And there are 4! arrangements of the four "letters".

    But in each arrangement, the 5 consonants can be ordered in 5! ways.

    Therefore, there are: . 4! \times 5! \:=\:24 \times 120 \:=\:2,\!880 arrangements
    . . with the consonants together.



    d) O and S are not together.

    We know there are 8! possible arrangements.

    Let's count the arrangement in which O and S are together.


    Duct-tape the O and S together.
    . . Then we have seven "letters" to arrange: . \boxed{OS},\,L,\,Y,\,M,\,P,\,I,\,C

    And there are 7! arrangements of the seven "letters."

    But in each arrangement, the OS could be ordered SO.

    Hence, there are: . 2 \times 7! arrangements with O and S together.


    Therefore, there are: . 8! - 2\!\cdot\!7! \:=\:40,320 - 10,080 \:=\:30,240 arrangements
    . . with O and S nonadjacent.

    Hello Soroban I appreciate all the help you have given me, although I am having trouble understanding parts of the answers.

    In c) why is it not 5!*3! because there are 3 vowels and 5 consonants?
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by william View Post
    Hello Soroban I appreciate all the help you have given me, although I am having trouble understanding parts of the answers.

    In c) why is it not 5!*3! because there are 3 vowels and 5 consonants?
    there are 4 objects that we are permuting. the 5 consonants in one block is treated as one object and then the other 3 vowels as 3 separate objects. there are 4! ways to arrange these. now, for each of those ways, there are 5! ways to arrange the 5 consonants within the block. hence 5!*4!
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