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Math Help - math question - inclusion exclusion principle

  1. #1
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    math question - inclusion exclusion principle

    x1+x2+x3+x4+x5=32
    3 of the variable are naturals and even
    2 of the variable are naturals and odd
    nobody of the variables equals to 1 or 0
    get the final solution

    i think i need to use the inclusion exclusion principle
    someone can help me with this please
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  2. #2
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    Quote Originally Posted by tukilala View Post
    x1+x2+x3+x4+x5=32
    3 of the variable are naturals and even
    2 of the variable are naturals and odd
    nobody of the variables equals to 1 or 0
    get the final solution

    i think i need to use the inclusion exclusion principle
    someone can help me with this please
    tukilala,

    I don't understand your problem statement.

    What do you mean by "get the final solution"? Do you mean "Find the number of solutions to the equation"?
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  3. #3
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    Quote Originally Posted by tukilala View Post
    x1+x2+x3+x4+x5=32
    3 of the variable are naturals and even
    2 of the variable are naturals and odd
    nobody of the variables equals to 1 or 0
    get the final solution

    i think i need to use the inclusion exclusion principle
    someone can help me with this please
    Let a,b,c,d,e be natural numbers.

    Let  2a = x_1
     2b = x_2
     2c = x_3
     2d+1 = x_4
     2e+1 = x_5

    Then your equation can be expressed:

     2a+2b+2c +2d+1+2e+1 = 32

     2a+2b+2c +2d+2e + 2 = 32

     2a+2b+2c +2d+2e  = 30

     2(a+b+c +d+e)  = 30

     a+b+c +d+e  = 15

    There are many solutions. To get one, just try some numbers. As long as they're natural, it doesn't matter.

    Since none are 0 or 1, I'd just put all 2s until the last one, and then see what you need to make up 15.

     2+2+2+2+7 = 15

    Hence
     a = 2
     b = 2
     c = 2
     d = 2
     e = 7

      x_1 = 4
     x_2 = 4
      x_3 = 4
      x_4 = 5
      x_5 = 15

    Is a solution.
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  4. #4
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    i dont understand somthing...

    i dont understand why u deside that
    2a=x1
    2b=x2
    2c=x3
    2d+1=x4
    2e+1=x5

    you dont know wich of the variable is odd and wich is even,
    you just know that there are 3 even and 2 odd
    so i think that you cant decide that x4 and x5 are odd by give them the value 2d+1 and 2e+1

    am i right? of not? why?
    please give me more ditails...
    thanks
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  5. #5
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    Quote Originally Posted by tukilala View Post
    i dont understand why u deside that
    2a=x1
    2b=x2
    2c=x3
    2d+1=x4
    2e+1=x5

    you dont know wich of the variable is odd and wich is even,
    you just know that there are 3 even and 2 odd
    so i think that you cant decide that x4 and x5 are odd by give them the value 2d+1 and 2e+1

    am i right? of not? why?
    please give me more ditails...
    thanks
    Yes.

    Any odd natural number can be expressed in the form 2k+1, where k is any natural number. Any even natural number can be expressed in the form 2k, where k is any natural number. That is why i choose these.
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  6. #6
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    but if x1 is odd? so x1 cant be equal to 2a
    my point is that you dont know wich number is odd and wich is even
    so how you decide that x1 is even and x5 is odd??????
    so i think something missing here, no?
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  7. #7
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    Quote Originally Posted by tukilala View Post
    but if x1 is odd? so x1 cant be equal to 2a
    my point is that you dont know wich number is odd and wich is even
    so how you decide that x1 is even and x5 is odd??????
    so i think something missing here, no?
    It doesn't matter what you decide.

    The question doesn't specify WHICH variables are odd or even, it just says that 3 are even, and 2 are odd. You are at liberty to decide which ones you want to make.

    The fact of the matter is that there are hundreds of different solutions to the equation under the given requirements. It doesn't really matter what one you choose.
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  8. #8
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    Quote Originally Posted by Mush View Post
    Let a,b,c,d,e be natural numbers.
    Let  2a = x_1 ,  2b = x_2 ,  2c = x_3
     2d+1 = x_4 ,  2e+1 = x_5
    So  a+b+c +d+e  = 15
    Mush’s solution is interesting and solves the problem if we specify that x_1 ,\,x_2 ,\;\& \,x_3 are the even variables and the others are odd. However, there are more solutions that Mush seems to indicate.
    For example: a = 1,\,b = 3,\,c = 4,\,d = 6\;\& \,e = 1 gives x_1  = 2,\,x_2  = 6,\,x_3  = 8,\,x_4  = 13\;\& \,x_5  = 3 which also works.

    So we must count the number of solutions to  a+b+c +d+e  = 15 in the positive integers (each variable at least 1).
    That number is {{10+5-1}\choose {10}}.
    As has been pointed out, we do not know which of the variables the odd ones are.
    So we to account for that consider that the two odds may be in {5\choose 2} places.
    The total then is {{10+5-1}\choose {10}} {5\choose 2}
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  9. #9
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    but why u did: (10+5-1)C(10)
    there are 15 digits
    so it is not sepose to be (15+5-1)C(15)
    ????
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  10. #10
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    Do you have a textbook that contains this material?

    Placing N identical objects into k different cells can be done in {{N+k-1} \choose {N}} ways.
    However, some of the cells may be empty.
    Like  a = 1,\,b = 0,\,c = 8,\,d = 6\;\& \,e = 0 is a solution to  a+b+c +d+e  = 15 in the nonnegative integers.

    Placing N identical objects into k different cells, with no empty cell, can be done in {{N-1} \choose {N-k}} ways.
    The solutions to  a+b+c +d+e  = 15 in the positive integers number {{15-1} \choose {15-5}}.
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