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Math Help - Number of combinations

  1. #1
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    Number of combinations

    Given you have 6 questions.
    Each question can have marks between 0 to 10
    The following question must be less or equal to the previous question.

    How many combination of set of marks is possible?
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  2. #2
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    Re: Number of combinations

    Quote Originally Posted by BabyMilo View Post
    Given you have 6 questions.
    Each question can have marks between 0 to 10
    The following question must be less or equal to the previous question.
    How many combination of set of marks is possible?
    I assume that you mean the marks are from 0 to 10, eleven scores in all.
    The back-of-the-book will tell you that the answer is \dbinom{16}{6}~.
    You tell us why and how.
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  3. #3
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    Re: Number of combinations

    First time posting and this is my solution, hope it right.

    Case 1: 6 questions have different number of marks.
    combination of set will be (9*8*7*6*5*4)/(6!)
    Explain: the first question has 9 choices of number of marks. the second one has 8 choices. the third one has 7 choices. the fourth one has 6 choices. the fifth one has 5 choices and the last one has 4 choices. and with condition that the following question must be less or equal to the previous question, so we have to divided by 6! because with 6 different numbers (excluded 0), we have 6! cases to put them together but there is only one case that suit the condition.

    Case 2: there are 2 questions have the same number of marks.
    let say combine these 2 questions to one question. so we have five questions now.
    therefore, the combination of set in this case will be (9*8*7*6*5)/(5!)

    Case 3: 3 questions have the same number of marks. combination of set: (9*8*7*6)/(4!)

    Case 4: 4 questions have the same number of marks. combination of set : (9*8*7)/(3!)

    case 5: 5 questions have the same number of marks. combination of set : (9*8)/(2!)

    case 5: 6 questions have the same number of marks. combination of set : 9

    by adding up all 6 case together we will get the solution: 465
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  4. #4
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    Re: Number of combinations

    Quote Originally Posted by tuanvu815 View Post
    First time posting and this is my solution, hope it right.

    Case 1: 6 questions have different number of marks.
    combination of set will be (9*8*7*6*5*4)/(6!)

    by adding up all 6 case together we will get the solution: 465
    @tuanvu815, how are you reading the quesrtion?
    The scores are non-increasing.
    This could be one paper: 9,9,7,6,6,6
    We cannot have a paper marked 4,5,6,2,1,0. Scores cannot increase.
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  5. #5
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    Re: Number of combinations

    I dont know why.
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  6. #6
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    Re: Number of combinations

    Quote Originally Posted by BabyMilo View Post
    I dont know why.
    At least tell us if the reading of the question in reply #4 is correct.
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  7. #7
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    Re: Number of combinations

    Quote Originally Posted by Plato View Post
    At least tell us if the reading of the question in reply #4 is correct.
    yes, its correct.

    your answer in #2 is correct.
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  8. #8
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    Re: Number of combinations

    Quote Originally Posted by BabyMilo View Post
    yes, its correct. your answer in #2 is correct.
    I will explain it to you using an equivalent example which is easy to understand.
    Consider the first eleven letters of the English alphabet: [TEX]\{A,B,C,D,E,F,G,H,I,J,K\}/TEX].
    Select any multi-set of six. (multi-sets allow repeating members)
    Say we select [E,K,K,A,I,I]
    There are \dbinom{16}{6} such possible selections.
    This string can be rearranged in \dfrac{6!}{2!\cdot2!} ways.
    BUT only of those arrangements is in alphabetical order.

    If the question were "how many six letter strings using the first eleven letters( repeats allowed) have there letters in alphabetical order?" The answer is \dbinom{16}{6},

    That is the same as the original question.
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  9. #9
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    Re: Number of combinations

    Quote Originally Posted by Plato View Post
    I will explain it to you using an equivalent example which is easy to understand.
    Consider the first eleven letters of the English alphabet: [TEX]\{A,B,C,D,E,F,G,H,I,J,K\}/TEX].
    Select any multi-set of six. (multi-sets allow repeating members)
    Say we select [E,K,K,A,I,I]
    There are \dbinom{16}{6} such possible selections.
    This string can be rearranged in \dfrac{6!}{2!\cdot2!} ways.
    BUT only of those arrangements is in alphabetical order.

    If the question were "how many six letter strings using the first eleven letters( repeats allowed) have there letters in alphabetical order?" The answer is \dbinom{16}{6},

    That is the same as the original question.
    i still dont see where the 16 is coming from.

    thanks for your help.
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  10. #10
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    Re: Number of combinations

    thanks Plato for showing problems in my answer. when i solve it, i didnt think about 9,9,7,6,6,6. lol
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    Re: Number of combinations

    Quote Originally Posted by BabyMilo View Post
    i still dont see where the 16 is coming from.
    If we select N items from K different kinds (repeats allowd) there are
    \dbinom{N+K-1}{N} possible multi-sets.
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  12. #12
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    Re: Number of combinations

    Never learn this.
    Is this a formula for all similar problem?
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  13. #13
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    Re: Number of combinations

    Quote Originally Posted by BabyMilo View Post
    Never learn this.
    Is this a formula for all similar problem?
    Your course has not studied multisets???
    Thanks from BabyMilo
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