Find the number of triples $(A, B, C)$ of subsets of $[n]$ such that at least one of $A cap B$, $A cap C$, or...












1












$begingroup$


Find the number of triples $(A, B, C)$ of subsets of $[n]$ such that at least one of $A cap B$, $A cap C$, or $B cap C$ is empty



I got $6^ncdot 3-3cdot 5^n+4^n$, not sure if this is correct.

Please help me on this.






combinatorics







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    $begingroup$
    Please explain how you got your answer. It is much easier to check work than to redo it. It is correct for $n=1$.
    $endgroup$
    – Ross Millikan
    Dec 16 '18 at 17:42
















1












$begingroup$


Find the number of triples $(A, B, C)$ of subsets of $[n]$ such that at least one of $A cap B$, $A cap C$, or $B cap C$ is empty



I got $6^ncdot 3-3cdot 5^n+4^n$, not sure if this is correct.

Please help me on this.






combinatorics







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Please explain how you got your answer. It is much easier to check work than to redo it. It is correct for $n=1$.
    $endgroup$
    – Ross Millikan
    Dec 16 '18 at 17:42














1












1








1





$begingroup$


Find the number of triples $(A, B, C)$ of subsets of $[n]$ such that at least one of $A cap B$, $A cap C$, or $B cap C$ is empty



I got $6^ncdot 3-3cdot 5^n+4^n$, not sure if this is correct.

Please help me on this.






combinatorics







share|cite|improve this question











$endgroup$




Find the number of triples $(A, B, C)$ of subsets of $[n]$ such that at least one of $A cap B$, $A cap C$, or $B cap C$ is empty



I got $6^ncdot 3-3cdot 5^n+4^n$, not sure if this is correct.

Please help me on this.






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 16 '18 at 20:10









t.ysn

1397




1397










asked Dec 16 '18 at 17:08









MojicaMojica

202




202








  • 3




    $begingroup$
    Please explain how you got your answer. It is much easier to check work than to redo it. It is correct for $n=1$.
    $endgroup$
    – Ross Millikan
    Dec 16 '18 at 17:42














  • 3




    $begingroup$
    Please explain how you got your answer. It is much easier to check work than to redo it. It is correct for $n=1$.
    $endgroup$
    – Ross Millikan
    Dec 16 '18 at 17:42








3




3




$begingroup$
Please explain how you got your answer. It is much easier to check work than to redo it. It is correct for $n=1$.
$endgroup$
– Ross Millikan
Dec 16 '18 at 17:42




$begingroup$
Please explain how you got your answer. It is much easier to check work than to redo it. It is correct for $n=1$.
$endgroup$
– Ross Millikan
Dec 16 '18 at 17:42










1 Answer
1






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$begingroup$

I assume $[n]={1,2,ldots,n}$. For an arbitrary triples $(A,B,C)$ (not requiring anything), each $kin [n]$ can stay in one of the following list of sets:




  • $[n]-(Acup Bcup C)$

  • $A-(Bcup C)$

  • $B-(Ccup A)$

  • $C-(Acup B)$

  • $Bcap C-Acap Bcap C$

  • $Ccap A -Acap Bcap C$

  • $Acap B-Acap Bcap C$


  • $Acap Bcap C$.


To meet your requirement, $k$ cannot stay in $Acap Bcap C$ and must omit one of the sets $Bcap C$, $Ccap A$, and $Acap B$.



Let $n_A$, $n_B$, and $n_C$ denote the number of triples $(A,B,C)$ s.t. each $kin [n]$ does not stay in $Bcap C$, $Ccap A$, and $Acap B$, respectively. To count $n_A$, each $k$ has only six choices from the list of sets above. That is, we have $n_A=6^n$, and likewise, $n_B=n_C=6^n$.



Now, write $n_{A,B}$ for the number of triples $(A,B,C)$ s.t. each $kin[n]$ does not stay in $Bcap C$ and $Ccap A$. Define $n_{B,C}$ and $n_{C,A}$ similarly. To count $n_{A,B}$, each $k$ has only $5$ choices from the list of sets above. That is, $n_{A,B}=5^n$, and likewise, $n_{B,C}=n_{C,A}=5^n$.



Finally, let $n_{A,B,C}$ be the number of triples $(A,B,C)$ s.t. each $[k]$ stays in none of $Bcap C$, $Ccap A$, and $Acap B$. Then each $k$ has only $4$ possible places to be in the list above. So $n_{A,B,C}=4^n$. By PIE the answer is
$$(n_A+n_B+n_C)-(n_{A,B}+n_{B,C}+n_{C,A})+n_{A,B,C}=3cdot 6^n-3cdot 5^n+4^n.$$
Your answer is correct.






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    $begingroup$

    I assume $[n]={1,2,ldots,n}$. For an arbitrary triples $(A,B,C)$ (not requiring anything), each $kin [n]$ can stay in one of the following list of sets:




    • $[n]-(Acup Bcup C)$

    • $A-(Bcup C)$

    • $B-(Ccup A)$

    • $C-(Acup B)$

    • $Bcap C-Acap Bcap C$

    • $Ccap A -Acap Bcap C$

    • $Acap B-Acap Bcap C$


    • $Acap Bcap C$.


    To meet your requirement, $k$ cannot stay in $Acap Bcap C$ and must omit one of the sets $Bcap C$, $Ccap A$, and $Acap B$.



    Let $n_A$, $n_B$, and $n_C$ denote the number of triples $(A,B,C)$ s.t. each $kin [n]$ does not stay in $Bcap C$, $Ccap A$, and $Acap B$, respectively. To count $n_A$, each $k$ has only six choices from the list of sets above. That is, we have $n_A=6^n$, and likewise, $n_B=n_C=6^n$.



    Now, write $n_{A,B}$ for the number of triples $(A,B,C)$ s.t. each $kin[n]$ does not stay in $Bcap C$ and $Ccap A$. Define $n_{B,C}$ and $n_{C,A}$ similarly. To count $n_{A,B}$, each $k$ has only $5$ choices from the list of sets above. That is, $n_{A,B}=5^n$, and likewise, $n_{B,C}=n_{C,A}=5^n$.



    Finally, let $n_{A,B,C}$ be the number of triples $(A,B,C)$ s.t. each $[k]$ stays in none of $Bcap C$, $Ccap A$, and $Acap B$. Then each $k$ has only $4$ possible places to be in the list above. So $n_{A,B,C}=4^n$. By PIE the answer is
    $$(n_A+n_B+n_C)-(n_{A,B}+n_{B,C}+n_{C,A})+n_{A,B,C}=3cdot 6^n-3cdot 5^n+4^n.$$
    Your answer is correct.






    share|cite|improve this answer









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      0












      $begingroup$

      I assume $[n]={1,2,ldots,n}$. For an arbitrary triples $(A,B,C)$ (not requiring anything), each $kin [n]$ can stay in one of the following list of sets:




      • $[n]-(Acup Bcup C)$

      • $A-(Bcup C)$

      • $B-(Ccup A)$

      • $C-(Acup B)$

      • $Bcap C-Acap Bcap C$

      • $Ccap A -Acap Bcap C$

      • $Acap B-Acap Bcap C$


      • $Acap Bcap C$.


      To meet your requirement, $k$ cannot stay in $Acap Bcap C$ and must omit one of the sets $Bcap C$, $Ccap A$, and $Acap B$.



      Let $n_A$, $n_B$, and $n_C$ denote the number of triples $(A,B,C)$ s.t. each $kin [n]$ does not stay in $Bcap C$, $Ccap A$, and $Acap B$, respectively. To count $n_A$, each $k$ has only six choices from the list of sets above. That is, we have $n_A=6^n$, and likewise, $n_B=n_C=6^n$.



      Now, write $n_{A,B}$ for the number of triples $(A,B,C)$ s.t. each $kin[n]$ does not stay in $Bcap C$ and $Ccap A$. Define $n_{B,C}$ and $n_{C,A}$ similarly. To count $n_{A,B}$, each $k$ has only $5$ choices from the list of sets above. That is, $n_{A,B}=5^n$, and likewise, $n_{B,C}=n_{C,A}=5^n$.



      Finally, let $n_{A,B,C}$ be the number of triples $(A,B,C)$ s.t. each $[k]$ stays in none of $Bcap C$, $Ccap A$, and $Acap B$. Then each $k$ has only $4$ possible places to be in the list above. So $n_{A,B,C}=4^n$. By PIE the answer is
      $$(n_A+n_B+n_C)-(n_{A,B}+n_{B,C}+n_{C,A})+n_{A,B,C}=3cdot 6^n-3cdot 5^n+4^n.$$
      Your answer is correct.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I assume $[n]={1,2,ldots,n}$. For an arbitrary triples $(A,B,C)$ (not requiring anything), each $kin [n]$ can stay in one of the following list of sets:




        • $[n]-(Acup Bcup C)$

        • $A-(Bcup C)$

        • $B-(Ccup A)$

        • $C-(Acup B)$

        • $Bcap C-Acap Bcap C$

        • $Ccap A -Acap Bcap C$

        • $Acap B-Acap Bcap C$


        • $Acap Bcap C$.


        To meet your requirement, $k$ cannot stay in $Acap Bcap C$ and must omit one of the sets $Bcap C$, $Ccap A$, and $Acap B$.



        Let $n_A$, $n_B$, and $n_C$ denote the number of triples $(A,B,C)$ s.t. each $kin [n]$ does not stay in $Bcap C$, $Ccap A$, and $Acap B$, respectively. To count $n_A$, each $k$ has only six choices from the list of sets above. That is, we have $n_A=6^n$, and likewise, $n_B=n_C=6^n$.



        Now, write $n_{A,B}$ for the number of triples $(A,B,C)$ s.t. each $kin[n]$ does not stay in $Bcap C$ and $Ccap A$. Define $n_{B,C}$ and $n_{C,A}$ similarly. To count $n_{A,B}$, each $k$ has only $5$ choices from the list of sets above. That is, $n_{A,B}=5^n$, and likewise, $n_{B,C}=n_{C,A}=5^n$.



        Finally, let $n_{A,B,C}$ be the number of triples $(A,B,C)$ s.t. each $[k]$ stays in none of $Bcap C$, $Ccap A$, and $Acap B$. Then each $k$ has only $4$ possible places to be in the list above. So $n_{A,B,C}=4^n$. By PIE the answer is
        $$(n_A+n_B+n_C)-(n_{A,B}+n_{B,C}+n_{C,A})+n_{A,B,C}=3cdot 6^n-3cdot 5^n+4^n.$$
        Your answer is correct.






        share|cite|improve this answer









        $endgroup$



        I assume $[n]={1,2,ldots,n}$. For an arbitrary triples $(A,B,C)$ (not requiring anything), each $kin [n]$ can stay in one of the following list of sets:




        • $[n]-(Acup Bcup C)$

        • $A-(Bcup C)$

        • $B-(Ccup A)$

        • $C-(Acup B)$

        • $Bcap C-Acap Bcap C$

        • $Ccap A -Acap Bcap C$

        • $Acap B-Acap Bcap C$


        • $Acap Bcap C$.


        To meet your requirement, $k$ cannot stay in $Acap Bcap C$ and must omit one of the sets $Bcap C$, $Ccap A$, and $Acap B$.



        Let $n_A$, $n_B$, and $n_C$ denote the number of triples $(A,B,C)$ s.t. each $kin [n]$ does not stay in $Bcap C$, $Ccap A$, and $Acap B$, respectively. To count $n_A$, each $k$ has only six choices from the list of sets above. That is, we have $n_A=6^n$, and likewise, $n_B=n_C=6^n$.



        Now, write $n_{A,B}$ for the number of triples $(A,B,C)$ s.t. each $kin[n]$ does not stay in $Bcap C$ and $Ccap A$. Define $n_{B,C}$ and $n_{C,A}$ similarly. To count $n_{A,B}$, each $k$ has only $5$ choices from the list of sets above. That is, $n_{A,B}=5^n$, and likewise, $n_{B,C}=n_{C,A}=5^n$.



        Finally, let $n_{A,B,C}$ be the number of triples $(A,B,C)$ s.t. each $[k]$ stays in none of $Bcap C$, $Ccap A$, and $Acap B$. Then each $k$ has only $4$ possible places to be in the list above. So $n_{A,B,C}=4^n$. By PIE the answer is
        $$(n_A+n_B+n_C)-(n_{A,B}+n_{B,C}+n_{C,A})+n_{A,B,C}=3cdot 6^n-3cdot 5^n+4^n.$$
        Your answer is correct.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 20:48







        user614671





































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