Math Problem....

This is a discussion on Math Problem.... within the A Brief History of Cprogramming.com forums, part of the Community Boards category; Hi everybody ... I just came back from my trip... it is very nice to see all these answers... Thanx ...

  1. #16
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    OOOOh My God...

    Hi everybody ...

    I just came back from my trip... it is very nice to see all these answers... Thanx for everybody ...

    I was in Getty Musiam , and I advice you all to visit it http://www.getty.edu it was a nice long tirp...

    I will read your answers now, as I will add some more tomorrow , thanx for everybody.

    I hope that you like this problem. (I can get you more if you want.)

    thanx again
    C++
    The best

  2. #17
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    Hi everybody

    Hmm, it was some time ago I read discrete mathematics... but I'll try .

    The 5 pencils can only be distributed in 1 way since they are identical, and have the same amount as the students.

    The placement of the eight books can be split into 3 parts, as shown in the figure.

    In part 1, the books can be distributed in 8! ways. But since the order of the books isn't important in column 5, you must divide by 4! so you get 8! / 4!. Also note that the fifth column can be in any of the five students, so multiply that by 5.
    You get (8! / 4!) * 5.

    In part 2, the books can be distributed in 8! ways. But the order of the books isn't important in the two last columns so you have to divide by 2! and 3!. You get 8! / (2! * 3!). The last two columns can be in any of the five students, so you have to multiply by 5 * 4 (the last column can be in any of the 5, the second last can be in any of the 5 minus the one the last one is in, giving 4).
    You get (8! / (2! * 3!)) * (5 * 4).

    In part 3, the books can be distributed in 8! ways. But the order of the books isn't important in the three last columns, so you have to divide by 2! three times. You get (8! / (2! * 2! * 2!)). The last three columns can be in any of the five students, so you have to multiply by 5 * 4 * 3.
    You get (8! / (2! * 2! * 2!)) * (5 * 4 * 3).

    Then sum all those:

    (8! / 4!) * 5 +
    (8! / (2! * 3!)) * (5 * 4) +
    (8! / (2! * 2! * 2!)) * (5 * 4 * 3)

    You might want to shorten that expression:
    8! * ((5 / 24) + (10 / 6) + (15 / 2))

    And the final result:
    378000
    Your Solution is realy right , but in part three you need to div by 3! , because the problem in part III is selection , but in the first 2 part is P arranging ...

    so the last answere would be like this
    (8! / 4!) * 5 +
    (8! / (2! * 3!)) * (5 * 4) +
    (8! / (2! * 2! * 2!)*3!) * (5 * 4 * 3)

    and that is it ....

    thanx again.
    C++
    The best

  3. #18
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    I do the same thing in all 3 cases, so I don't see why you would also divide by 3! in the last case ???

    8! is the 8 distributed books

    (2! * 2! * 2!) is divided cause in each 2-book pile, the books can be arranged in 2 ways which doesn't matter, therefore divide by 2 three times.

    (5 * 4 * 3) is the distribution of the book piles. The first pile can be placed in any of the five students. The second pile can be placed in any of the 4 remaining students. The last pile can be placed in any of the 3 remaining students.

    Why do you divide by 3! ???
    MagosX.com

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  4. #19
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    ok let me show you my way...

    when I solve it in the class , I solve it in this way...

    we have 5 students ... e1 , e2, ... e5

    so we have
    e1 + e2 + e3+ e4+ e5 = 8

    ei is one one the student 1 =<ei =<5 and i>=1


    g(x) = ( x + xpwr2 + .... + xpwr8)pwr5
    = (x+( x pwr2)/2! + .... (x pwr8)/8!) pwr5
    = ( ( e pwrx) -1) pwr5
    = [e pwr(5x)] - [c(5,1) e pwr 4x] + ... -.... +[pc(5,4) e pwr x ]-1
    = c(5,0) SEGMA 5 (x pwr 8) / 8! - c(5,1) SEGMA 4 (x pwr 4 )/4! +... -..... -1
    = {SEGMA [ 5 pwr 8 - C(5,1) 4 pwr 8 + C(5,2) 3 pwr 8 - C(5,3) 2 pwr 8 + C(5,4) 1 pwr 8 ] x pwr8 / 8! }-1

    A8 = 5 pwr8 - C(5,1) 4 pwr 8 + C(5,2) 3 pwr 8 - C(5,3) 2 pwr 8 + C(5,4) 1 pwr 8

    count the answer, and once you have it , look at the last answer and devid it by 3! you will find that the resuts sound be the same.

    as I told you ... the way that you solve it is more programming than what I did ( which is great , because you are a good programmer and you have a great future in that ) but you forgot that in the part III you need to select not to destribute ... and that is why you need to devid by 3!.

    I hope that you understand my poor English.

    Again. I would like to say ... that your way is a very computerizad way and ( I - should use it because I am a programmer - ) but you need to devid by 3! for the last part to get the right answer for your solution.

    thanx again
    C++
    The best

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