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Math Help - Need these within 1 hour !!!

  1. #1
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    Need these within 1 hour !!!

    1. In how many way could 15 different books be divided equally among 3 people?
    Answer in the book = 756756

    2. The camera club has 5 members, and the mathematics club has 8. There is only one member common to both clubs. In how many ways could a committee of 4 people be formed with at least one member from each club?
    Answer in the book = 459

    3. There are 12 questions on an examination , and each student must answer 8 questions including at least 4 of the first 5 questions. How many different combinations of questions could a student choose to answer?
    Answer in the book = 210
    Please help me !!!
    Last edited by Faisal2007; October 6th 2008 at 02:17 PM.
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  2. #2
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    Hey there faisal,

    Because of time zones, I'm not sure if I'm too late for you...

    1. In how many way could 15 different books be divided equally among 3 people?
    Answer in the book = 756756

    15 books can be divided 15! ways = 1307674368000

    Even though the books are different, the order within the group is not important. ie. ABCDE = ADECB (both groups contain the same letters). Each group of 5 books can be divided in 5! ways.

    Hence, 15 different books are divided eually among 3 people

    \frac {15!}{5!*5!*5!} = \frac {15!}{5!^3} = 756756 ways.

    2. The camera club has 5 members, and the mathematics club has 8. There is only one member common to both clubs. In how many ways could a committee of 4 people be formed with at least one member from each club?
    Answer in the book = 459

    If you know what a venn diagram is, it might be helpful to draw problems like this using one...

    There are 12 members all together (5 + 8 - 1(the common member) = 12)

    These 12 members can be arranged in \frac {12!}{4!} = 495 ways.

    This number represents the total permutations for the group. Taking out the other groups which contain only members from one group:

    495 - \frac {4!}{4!} - \frac {7*6*5*4}{4!} = 459 ways.

    I think you've mixed up some information for your 3rd question

    Sorry about the delay - I'm on child-duties at home :P

    Trust that helps...
    Last edited by MakeANote; October 2nd 2008 at 03:42 PM.
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  3. #3
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    Fixed question number 3.

    Although the due date is over and i still havent gotten question # 2 and 3 i am DIEING to know the concept behind them.

    Thanks for the answers bro.

    I didnt quite get your answer number 2...
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  4. #4
    Member Henderson's Avatar
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    On #3:

    If a student does all of the first 5 problems, then he must find 3 more problems to do out of the 7 remaining problems available. There are \frac{7*6*5}{3!} = 35 ways to do this.

    If a student does only 4 of the first five, There are 5 ways to select those four problems (think of it as choosing a problem to skip). After those 4, he must find 4 more problems to do out of the 7 remaining problems available. There are \frac{7*6*5*4}{4!} = 35 ways to do this. Multiply the 5 ways to select the first four problems by the 35 ways to select the last four problems to get 175.




    175 ways to do just 4 of the first 5
    +35 ways to do all of the first 5

    210 ways to do 8 of the 12 problems
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