Find the cardinality of set ${(x,y,z):x,y,z in mathbb{Z}^+ wedge xyz=108}$












0












$begingroup$


Find the cardinality of set ${(x,y,z): x,y,ztext{ are positive integers and }xyz=108}$



my attempt:



I write all factors of $108$.



$$1,2,3,4,6,9,12,18,27,36,54,108.$$
First observe that $xyz $ has only two degree of freedom. So its enough to choose $x,y$ . If I choose $x=1$ there are $12$ ways of choosing $y$. If choose $x$ as $2$ there are $11$ ways ...and so on.



So total number of ways $=1+2+...12=78$.
But its give in my book that $60$ is answer. Where am I wrong?






combinatorics permutations combinations integers







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  • 1




    $begingroup$
    "If I choose $2$ there are $11$ ways of choosing $y$" Really? I only found $8$ ways: $1,2,3,6,9,18,27,54$. What are the other $3$ ways???
    $endgroup$
    – bof
    Jan 15 at 7:36










  • $begingroup$
    @bof oh...thanks
    $endgroup$
    – Cloud JR
    Jan 15 at 7:38
















0












$begingroup$


Find the cardinality of set ${(x,y,z): x,y,ztext{ are positive integers and }xyz=108}$



my attempt:



I write all factors of $108$.



$$1,2,3,4,6,9,12,18,27,36,54,108.$$
First observe that $xyz $ has only two degree of freedom. So its enough to choose $x,y$ . If I choose $x=1$ there are $12$ ways of choosing $y$. If choose $x$ as $2$ there are $11$ ways ...and so on.



So total number of ways $=1+2+...12=78$.
But its give in my book that $60$ is answer. Where am I wrong?






combinatorics permutations combinations integers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    "If I choose $2$ there are $11$ ways of choosing $y$" Really? I only found $8$ ways: $1,2,3,6,9,18,27,54$. What are the other $3$ ways???
    $endgroup$
    – bof
    Jan 15 at 7:36










  • $begingroup$
    @bof oh...thanks
    $endgroup$
    – Cloud JR
    Jan 15 at 7:38














0












0








0





$begingroup$


Find the cardinality of set ${(x,y,z): x,y,ztext{ are positive integers and }xyz=108}$



my attempt:



I write all factors of $108$.



$$1,2,3,4,6,9,12,18,27,36,54,108.$$
First observe that $xyz $ has only two degree of freedom. So its enough to choose $x,y$ . If I choose $x=1$ there are $12$ ways of choosing $y$. If choose $x$ as $2$ there are $11$ ways ...and so on.



So total number of ways $=1+2+...12=78$.
But its give in my book that $60$ is answer. Where am I wrong?






combinatorics permutations combinations integers







share|cite|improve this question











$endgroup$




Find the cardinality of set ${(x,y,z): x,y,ztext{ are positive integers and }xyz=108}$



my attempt:



I write all factors of $108$.



$$1,2,3,4,6,9,12,18,27,36,54,108.$$
First observe that $xyz $ has only two degree of freedom. So its enough to choose $x,y$ . If I choose $x=1$ there are $12$ ways of choosing $y$. If choose $x$ as $2$ there are $11$ ways ...and so on.



So total number of ways $=1+2+...12=78$.
But its give in my book that $60$ is answer. Where am I wrong?






combinatorics permutations combinations integers




combinatorics permutations combinations integers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 7:52









David G. Stork

11k41432




11k41432










asked Jan 15 at 6:59









Cloud JRCloud JR

909518




909518








  • 1




    $begingroup$
    "If I choose $2$ there are $11$ ways of choosing $y$" Really? I only found $8$ ways: $1,2,3,6,9,18,27,54$. What are the other $3$ ways???
    $endgroup$
    – bof
    Jan 15 at 7:36










  • $begingroup$
    @bof oh...thanks
    $endgroup$
    – Cloud JR
    Jan 15 at 7:38














  • 1




    $begingroup$
    "If I choose $2$ there are $11$ ways of choosing $y$" Really? I only found $8$ ways: $1,2,3,6,9,18,27,54$. What are the other $3$ ways???
    $endgroup$
    – bof
    Jan 15 at 7:36










  • $begingroup$
    @bof oh...thanks
    $endgroup$
    – Cloud JR
    Jan 15 at 7:38








1




1




$begingroup$
"If I choose $2$ there are $11$ ways of choosing $y$" Really? I only found $8$ ways: $1,2,3,6,9,18,27,54$. What are the other $3$ ways???
$endgroup$
– bof
Jan 15 at 7:36




$begingroup$
"If I choose $2$ there are $11$ ways of choosing $y$" Really? I only found $8$ ways: $1,2,3,6,9,18,27,54$. What are the other $3$ ways???
$endgroup$
– bof
Jan 15 at 7:36












$begingroup$
@bof oh...thanks
$endgroup$
– Cloud JR
Jan 15 at 7:38




$begingroup$
@bof oh...thanks
$endgroup$
– Cloud JR
Jan 15 at 7:38










2 Answers
2






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3












$begingroup$

$$(2^{j_1} 3^{k_1})(2^{j_2}3^{k_2})(2^{j_3}3^{k_3}) = 108 = 2^2 cdot 3^3$$



where ${ j_i, k_i} ge 0$ and by the Prime Factorization Theorem we have $j_1 + j_2 + j_3 = 2$ and $k_1 + k_2 + k_3 = 3$.



How many ways can you solve those equations, up to equivalent permutations of the factors? $12$



Start with the first (smallest) factor $(2^0 3^0)$ and increment others for the first "row"; then $(2^1 3^0)$ and increment others for the second "row"; then $(2^0 3^1)$ and so on... for the third "row." There is no fourth "row"...



Here are the sets, each in ascending order, themselves ordered:




  • {1, 1, 108}, {1, 2, 54}, {1, 3, 36}, {1, 4, 27}, {1, 6, 18}, {1, 9, 12}

  • {2, 2, 27}, {2, 3, 18}, {2, 6, 9}

  • {3, 3, 12}, {3, 4, 9}, {3, 6, 6}






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    $108=2^2 cdot 3^3$, so $x,y,z$ must be of the form $2^i3^j$ such that the sum of the $i$'s (where $0le ile 2$) is $2$ and the sum of the $j$'s (also integers starting from $0$) is $3$. How many solutions do $i_x+i_y+i_z = 2$ and $j_x+j_y+j_z = 3$ have in integers $ge 0$? All combinations of these solutions are the solutions you want.



    There are $binom{n+k-1}{k-1}$ non-negative integer solutions to $x_1+ ldots +x_k =n$ (see wikipedia, e.g.) and so for $3$ variables and $n=2$ we get $binom{2+3-1}{2} = binom{4}{2}=6$ and for $n=3$ we get $binom{5}{2} = 10$. So combined we get $6 times 10=60$ solutions for these independent equations.






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      3












      $begingroup$

      $$(2^{j_1} 3^{k_1})(2^{j_2}3^{k_2})(2^{j_3}3^{k_3}) = 108 = 2^2 cdot 3^3$$



      where ${ j_i, k_i} ge 0$ and by the Prime Factorization Theorem we have $j_1 + j_2 + j_3 = 2$ and $k_1 + k_2 + k_3 = 3$.



      How many ways can you solve those equations, up to equivalent permutations of the factors? $12$



      Start with the first (smallest) factor $(2^0 3^0)$ and increment others for the first "row"; then $(2^1 3^0)$ and increment others for the second "row"; then $(2^0 3^1)$ and so on... for the third "row." There is no fourth "row"...



      Here are the sets, each in ascending order, themselves ordered:




      • {1, 1, 108}, {1, 2, 54}, {1, 3, 36}, {1, 4, 27}, {1, 6, 18}, {1, 9, 12}

      • {2, 2, 27}, {2, 3, 18}, {2, 6, 9}

      • {3, 3, 12}, {3, 4, 9}, {3, 6, 6}






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        $$(2^{j_1} 3^{k_1})(2^{j_2}3^{k_2})(2^{j_3}3^{k_3}) = 108 = 2^2 cdot 3^3$$



        where ${ j_i, k_i} ge 0$ and by the Prime Factorization Theorem we have $j_1 + j_2 + j_3 = 2$ and $k_1 + k_2 + k_3 = 3$.



        How many ways can you solve those equations, up to equivalent permutations of the factors? $12$



        Start with the first (smallest) factor $(2^0 3^0)$ and increment others for the first "row"; then $(2^1 3^0)$ and increment others for the second "row"; then $(2^0 3^1)$ and so on... for the third "row." There is no fourth "row"...



        Here are the sets, each in ascending order, themselves ordered:




        • {1, 1, 108}, {1, 2, 54}, {1, 3, 36}, {1, 4, 27}, {1, 6, 18}, {1, 9, 12}

        • {2, 2, 27}, {2, 3, 18}, {2, 6, 9}

        • {3, 3, 12}, {3, 4, 9}, {3, 6, 6}






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          $$(2^{j_1} 3^{k_1})(2^{j_2}3^{k_2})(2^{j_3}3^{k_3}) = 108 = 2^2 cdot 3^3$$



          where ${ j_i, k_i} ge 0$ and by the Prime Factorization Theorem we have $j_1 + j_2 + j_3 = 2$ and $k_1 + k_2 + k_3 = 3$.



          How many ways can you solve those equations, up to equivalent permutations of the factors? $12$



          Start with the first (smallest) factor $(2^0 3^0)$ and increment others for the first "row"; then $(2^1 3^0)$ and increment others for the second "row"; then $(2^0 3^1)$ and so on... for the third "row." There is no fourth "row"...



          Here are the sets, each in ascending order, themselves ordered:




          • {1, 1, 108}, {1, 2, 54}, {1, 3, 36}, {1, 4, 27}, {1, 6, 18}, {1, 9, 12}

          • {2, 2, 27}, {2, 3, 18}, {2, 6, 9}

          • {3, 3, 12}, {3, 4, 9}, {3, 6, 6}






          share|cite|improve this answer











          $endgroup$



          $$(2^{j_1} 3^{k_1})(2^{j_2}3^{k_2})(2^{j_3}3^{k_3}) = 108 = 2^2 cdot 3^3$$



          where ${ j_i, k_i} ge 0$ and by the Prime Factorization Theorem we have $j_1 + j_2 + j_3 = 2$ and $k_1 + k_2 + k_3 = 3$.



          How many ways can you solve those equations, up to equivalent permutations of the factors? $12$



          Start with the first (smallest) factor $(2^0 3^0)$ and increment others for the first "row"; then $(2^1 3^0)$ and increment others for the second "row"; then $(2^0 3^1)$ and so on... for the third "row." There is no fourth "row"...



          Here are the sets, each in ascending order, themselves ordered:




          • {1, 1, 108}, {1, 2, 54}, {1, 3, 36}, {1, 4, 27}, {1, 6, 18}, {1, 9, 12}

          • {2, 2, 27}, {2, 3, 18}, {2, 6, 9}

          • {3, 3, 12}, {3, 4, 9}, {3, 6, 6}







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 15 at 7:44

























          answered Jan 15 at 7:19









          David G. StorkDavid G. Stork

          11k41432




          11k41432























              3












              $begingroup$

              $108=2^2 cdot 3^3$, so $x,y,z$ must be of the form $2^i3^j$ such that the sum of the $i$'s (where $0le ile 2$) is $2$ and the sum of the $j$'s (also integers starting from $0$) is $3$. How many solutions do $i_x+i_y+i_z = 2$ and $j_x+j_y+j_z = 3$ have in integers $ge 0$? All combinations of these solutions are the solutions you want.



              There are $binom{n+k-1}{k-1}$ non-negative integer solutions to $x_1+ ldots +x_k =n$ (see wikipedia, e.g.) and so for $3$ variables and $n=2$ we get $binom{2+3-1}{2} = binom{4}{2}=6$ and for $n=3$ we get $binom{5}{2} = 10$. So combined we get $6 times 10=60$ solutions for these independent equations.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                $108=2^2 cdot 3^3$, so $x,y,z$ must be of the form $2^i3^j$ such that the sum of the $i$'s (where $0le ile 2$) is $2$ and the sum of the $j$'s (also integers starting from $0$) is $3$. How many solutions do $i_x+i_y+i_z = 2$ and $j_x+j_y+j_z = 3$ have in integers $ge 0$? All combinations of these solutions are the solutions you want.



                There are $binom{n+k-1}{k-1}$ non-negative integer solutions to $x_1+ ldots +x_k =n$ (see wikipedia, e.g.) and so for $3$ variables and $n=2$ we get $binom{2+3-1}{2} = binom{4}{2}=6$ and for $n=3$ we get $binom{5}{2} = 10$. So combined we get $6 times 10=60$ solutions for these independent equations.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  $108=2^2 cdot 3^3$, so $x,y,z$ must be of the form $2^i3^j$ such that the sum of the $i$'s (where $0le ile 2$) is $2$ and the sum of the $j$'s (also integers starting from $0$) is $3$. How many solutions do $i_x+i_y+i_z = 2$ and $j_x+j_y+j_z = 3$ have in integers $ge 0$? All combinations of these solutions are the solutions you want.



                  There are $binom{n+k-1}{k-1}$ non-negative integer solutions to $x_1+ ldots +x_k =n$ (see wikipedia, e.g.) and so for $3$ variables and $n=2$ we get $binom{2+3-1}{2} = binom{4}{2}=6$ and for $n=3$ we get $binom{5}{2} = 10$. So combined we get $6 times 10=60$ solutions for these independent equations.






                  share|cite|improve this answer











                  $endgroup$



                  $108=2^2 cdot 3^3$, so $x,y,z$ must be of the form $2^i3^j$ such that the sum of the $i$'s (where $0le ile 2$) is $2$ and the sum of the $j$'s (also integers starting from $0$) is $3$. How many solutions do $i_x+i_y+i_z = 2$ and $j_x+j_y+j_z = 3$ have in integers $ge 0$? All combinations of these solutions are the solutions you want.



                  There are $binom{n+k-1}{k-1}$ non-negative integer solutions to $x_1+ ldots +x_k =n$ (see wikipedia, e.g.) and so for $3$ variables and $n=2$ we get $binom{2+3-1}{2} = binom{4}{2}=6$ and for $n=3$ we get $binom{5}{2} = 10$. So combined we get $6 times 10=60$ solutions for these independent equations.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 15 at 18:17

























                  answered Jan 15 at 7:12









                  Henno BrandsmaHenno Brandsma

                  110k347116




                  110k347116






























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