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Math Help - Counting and Combinatorics

  1. #1
    Member billym's Avatar
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    Counting and Combinatorics

    Q: How many solutions are there to:

    x1 + x2 + x3 + x4 +x5 = 50 for integers xi >= 1

    What I've done:

    Rewrite as: (x1 - 2) + (x2 - 2) + (x3 - 2) + (x4 - 2) + (x5 - 2) = 40

    My answer: The number of solutions is C(40 + 5 - 1 , 40 ) = 135751

    The answer has been given as 130905. Any idea where I have gone wrong?
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  2. #2
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    Hello, billym!

    I don't agree with their answer . . .


    How many solutions are there to:

    . . x_1 + x_2 + x_3 + x_4 +x_5 \:=\: 50 . for integers x_i \geq 1

    The answer has been given as 130,905.

    Consider a 50-inch dowel ... marked in one-inch intervals.
    . . \square\square\square\square\square\square\square\  square\hdots\square\square

    There are 49 marks along the dowel.
    We will select 4 of them to make our cuts, dividing the dowel into 5 parts.

    There are: . _{49}C_4 \:=\:{49\choose4} \:=\:\frac{49!}{4!\,45!} \;=\; 211,\!876 choices.

    Therefore, there are 211,876 solutions.

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  3. #3
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    If each answer is a at least one, then we are dealing with xL1+xL2+xL3+xL4+xL5=45 having "put" a one into each variable.
    So the solution is \binom{45+5-1}{45}.

    Here is the thought experience.
    Think of putting a 1 into each of the five variables.
    That leaves 45 ones to place into five different cells: \binom{45+5-1}{45}.
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  4. #4
    Super Member Random Variable's Avatar
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    If you expand  (x+x^{2}+x^{3}+...)^{5} , the coefficient of the x^{50} term is 211876
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  5. #5
    Member billym's Avatar
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    Hmm...

    That was the first, first answer I had:

    Rewrite as: (x1 - 1) + (x2 - 1) + (x3 - 1) + (x4 - 1) + (x5 - 1) = 45

    Thus, C(45 + 5 - 1 , 40 ) = 211876


    Guess the answer sheet is wrong. I hate that.
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  6. #6
    Super Member Random Variable's Avatar
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    Your answer is correct if  x_{i} \ge 2
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