Results 1 to 14 of 14

Math Help - 2 hard question....help plz!!!!

  1. #1
    Newbie sonymd23's Avatar
    Joined
    Jul 2006
    Posts
    22

    2 hard question....help plz!!!!

    can any one do number #6,10 & 11......thx very much every one......and show ur steps thnx.....cant thank you enough

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    [QUOTE=sonymd23]can any one do number #6,10 & 11......thx very much every one......and show ur steps thnx.....cant thank you enough

    [qUOTE]

    I'm pretty sure you will find the answer to 6 by hunting around this site for
    what was said when these questions were previously posted.

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    I know #10 was answered before by Soroban...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    I think I found the post
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Quick
    I think I found the post
    When I look at this thread the old Tex does not render unless I edit a
    post (make no changes) then save. Is this what everyone else sees
    happen ?

    RonL

    Looking at the printable version also works

    This thread shows why I would prefer that the image file be uploaded here
    rather than hosted somewhere like PhotoBucket - in that thread the questions
    can no longer be seen as the file has been removed from PhotoBucket.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by CaptainBlack
    When I look at this thread the old Tex does not render unless I edit a
    post (make no changes) then save. Is this what everyone else sees
    happen ?

    RonL

    Looking at the printable version also works

    This thread shows why I would prefer that the image file be uploaded here
    rather than hosted somewhere like PhotoBucket - in that thread the questions
    can no longer be seen as the file has been removed from PhotoBucket.
    Yes, the LaTex is having some troubles, I wonder why? could it have anything to do with the fact you can click on it and get the code?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Try using a different internet brouswer.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie sonymd23's Avatar
    Joined
    Jul 2006
    Posts
    22
    but can any one do number 10?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by sonymd23
    but can any one do number 10?
    Quote Originally Posted by MIA
    10. A school dance has 10 volunteers. .Each dance requires 2 volunteers at the door,
    3 volunteers on the floor, and 5 floaters. .If two of the volunteers, John and Tom,
    can not work together, in how many ways can the volunteers be assigned?
    Quote Originally Posted by Soroban*
    There are: \binom{10}{2,3,5}\:=\:\frac{10!}{2!3!5!}\:=\;2520 ways to make the assignments.

    We will count the number of ways that John and Tom are together.
    Duct-tape them together; then we have nine "people" to arrange.
    . . \{JT,\;A,\;B,\;C,\;D,\,E,\;F,\;G,\:H\}

    If JT are at the door, the other 8 people can be assigned in
    . . \binom{8}{3,5} \:=\:\frac{8!}{3!5!}\:=\:56 ways.

    If JT are on the floor, the other 8 people can be assigned in
    . . \binom{8}{2,1,5}\:=\:\frac{8!}{2!1!5!}\:=\:126 ways.

    If JT are floaters, the other 8 people can be assigned in
    . . \binom{8}{2,3,3}\:=\:\frac{8!}{2!3!3!}\:=\:560 ways.

    Hence, there are: 56 + 126 + 560\,=\,742 ways that John and Tom are together.


    Therefore, there are: 2520 - 742 \,=\,1778 ways that they are not together.
    *this question was answered in Quick's hyperlink
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie sonymd23's Avatar
    Joined
    Jul 2006
    Posts
    22
    Quote Originally Posted by Quick
    *this question was answered in Quick's hyperlink

    however my teacher said thats incorrect......he wrote.....

    john on floor: 8c2 x 7c2 x 5c5
    " " to door : 8c1 x 8c3 x 5c5
    " " as floater : 8c4 x 5c2 x 3c3

    so which one is correct?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    Quote Originally Posted by sonymd23
    however my teacher said thats incorrect......he wrote.....

    john on floor: 8c2 x 7c2 x 5c5
    " " to door : 8c1 x 8c3 x 5c5
    " " as floater : 8c4 x 5c2 x 3c3

    so which one is correct?
    I think your teacher is correct.
    I have one problem in Soroban,s solution: Why he had used the expression \frac{10!}{5!3!2!}?Please clarify.

    Keep Smiling
    Malay
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,740
    Thanks
    645
    Hello, sonymd23!

    however my teacher said thats incorrect......he wrote.....

    John on floor: 8c2 x 7c2 x 5c5
    " " to door : 8c1 x 8c3 x 5c5
    " " as floater : 8c4 x 5c2 x 3c3
    Are you sure this what he wrote?

    Sorry, but I have no idea what your teacher is doing . . .

    None of the expressions make sense . . .


    He said "John on floor: _8C_2 \times\, _6C_2 \times\, _5C_5"
    . . Does anyone else agree that this is totally contrary to any kind of logic?

    It says: John is on the floor. .(Hence, Tom is not on the floor.)

    Then _8C_2: we choose two from a group of eight people . . . for what?
    . . Why pick two people? .What eight people?
    If John is already assigned to the Floor and Tom is not, there are ten people to choose from.

    Then _6C_2: we choose two from a group of six people.
    . . Again, why two people . . . what six people?
    Since John is already assigned and we've chosen 2 people (in the preceding),
    . . then there are nine people to choose from (8, if we exclude Tom).

    Finally _5C_5: take the remaining five people and place them anywhere?

    Someone has been sipping way too much coughsyrup . . .

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    The problem I had solved was different; it had ten people.

    This problem has twelve people:
    . . 2 at the door, 4 on the floor, 6 floaters.
    Again, John and Tom will not work together.

    I'll solve it both ways . . .

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    There are three possible assignments for John.

    [1] John is at the Door . . . and Tom is not.
    The other one person at the Door is chosen from the other 10 people.
    . . There are: _{10}C_1 = 10 ways.
    The remaining 10 people are assigned to the 4 Floors and 6 Floaters.
    . . There are \begin{pmatrix}10\\4,6\end{pmatrix} = 210 ways.
    Hence, there are: 10 \times 210 = \boxed{2100} ways that John can be at the Door.

    [2] John is on the Floor . . . and Tom is not.
    The other 3 people on the Floor are chosen from the other 10 people.
    . . There are: _}10}C_3 = 120 ways.
    The remaining 8 people are assigned to the 2 Doors and 6 Floaters.
    . . There are: \begin{pmatrix}8\\2,6\end{pmatrix} = 28 ways.
    Hence, there are: 120 \times 28 = \boxed{3360} ways that John can be on the Floor.

    [3] John is a Floater . . . and Tom is not.
    The other 5 Floaters are chosen from the other 10 people.
    . . There are: _{10}C_5 = 252 ways.
    The remaing 5 people are assigned to the 2 Doors and 4 Floors.
    . . There are: \begin{pmatrix}6\\2,4\end{pmatrix} = 15 ways.
    Hence, there are: 252 \times 15 = \boxed{3780} ways that John can be a Floater.

    Therefore, there are: . 2100 + 3360 + 3780 \:= 9240 ways to assign the positions.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    There are: \begin{pmatrix}12\\2,4,6\end{pmatrix} = 13,860 assignments with no restrictions.

    Let's count the ways that John and Tom are together.
    Duct-tape them together and we have 11 "people" to assign.
    There are three cases . . .

    JT is at the Door.
    Then there are \begin{pmatrix}10\\4,6\end{pmatrix} = 210 ways to assign the other ten people.

    JT is on the Floor.
    Then there are \begin{pmatrix}10\\2,2,6\end{pmatrix} = 1260 ways to assign the other ten people.

    JT is a Floater.
    Then there are \begin{pmatrix}10\\2,4,4\end{pmatrix} = 3150 ways to assign the other ten people.


    Hence, there are 210 + 1260 + 3150 = 4620 ways for John and Tom to serve together.


    Therefore, there are 13,860 - 4620 \,= 9240 ways that John and Tom do not serve together.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Who is right?
    I'm through explaining . . . It's your call . . .

    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    I thought the teacher's answer was for 10 volunteers...I am sorry.

    Why did you use Hello Soroban!
    <br />
\begin{pmatrix}12\\2,4,6\end{pmatrix} = 13,860<br />
    Please explain.
    I thought it was used when some of the things are alike.

    Keep Smiling
    Malay
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,740
    Thanks
    645
    Hello, Malay!

    Why did you use: \begin{pmatrix}12\\2,4,6\end{pmatrix} = 13,860

    Please explain.
    I thought it was used when some of the things are alike.

    You are correct!

    The twelve people are different (distinguisable), of course.

    But selecting two people for the Door is a combination: \begin{pmatrix}12\\2\end{pmatrix}
    . . The order in which the two people are chosen is not considered.

    Then we select 4 people for the Floor from the remaining 10 people: \begin{pmatrix}10\\4\end{pmatrix}
    . . Again, the order of the four names is not important.

    Finally, we select 6 people to be Floaters from the remaining 6 people: \begin{pmatrix}6\\6\end{pmatrix}
    . . And their order is ignored . . . so there is only one way.


    The total number of assignments is: . \begin{pmatrix}12\\2\end{pmatrix}\begin{pmatrix}10  \\4\end{pmatrix}\begin{pmatrix}6\\6\end{pmatrix}

    . . = \;\frac{12!}{2!\not{10!}}\cdot\frac{\not{10!}}{4! \not{6!}}\cdot\frac{\not{6!}}{6!\not{0!}} \;=\;\frac{12!}{2!\,4!\,6!} \;= \begin{pmatrix}12\\2,4,6\end{pmatrix}<br /> <br />

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. cos question hard
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: September 20th 2011, 01:37 PM
  2. hard question!
    Posted in the Geometry Forum
    Replies: 4
    Last Post: June 14th 2009, 03:31 AM
  3. Please Help, Hard Question for me.
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: January 28th 2009, 01:24 PM
  4. This question is so hard!! any help??
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 4th 2008, 11:59 AM
  5. Hard question
    Posted in the Statistics Forum
    Replies: 1
    Last Post: April 13th 2008, 05:38 PM

Search Tags


/mathhelpforum @mathhelpforum