'selfish' set to be a set which has its own cardinality (number of elements) as an element












4














Define a selfish set to be a set which has its own cardinality (number of elements) as an element. Find, with proof, the number of subsets of ${1, 2, ldots, n}$ which are minimal selfish sets, that is, selfish sets none of whose proper subsets is selfish.



My Attempt:
Assume $textbf{A}$ to be a selfish set. If the cardinality of $textbf{A}$ is $c$, then can $textbf{A}$ contain $1,2,3....c-1$. Definitely answer is no. because if it contains $k<c$ then deleting $c-k$ elements except $k$ from $textbf{A}$ gives a subset of k elements contradicting the fact that $textbf{A}$ is minimal selfish.
Thus $textbf{A}$ must contain elements greater than or equal to $c$. But how do I find the minimal selfish sets with order $c$?






combinatorics discrete-mathematics







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  • Reference note: This question, including the terminology, was problem B1 on the 1996 Putnam.
    – jmerry
    Dec 8 at 7:51
















4














Define a selfish set to be a set which has its own cardinality (number of elements) as an element. Find, with proof, the number of subsets of ${1, 2, ldots, n}$ which are minimal selfish sets, that is, selfish sets none of whose proper subsets is selfish.



My Attempt:
Assume $textbf{A}$ to be a selfish set. If the cardinality of $textbf{A}$ is $c$, then can $textbf{A}$ contain $1,2,3....c-1$. Definitely answer is no. because if it contains $k<c$ then deleting $c-k$ elements except $k$ from $textbf{A}$ gives a subset of k elements contradicting the fact that $textbf{A}$ is minimal selfish.
Thus $textbf{A}$ must contain elements greater than or equal to $c$. But how do I find the minimal selfish sets with order $c$?






combinatorics discrete-mathematics







share|cite|improve this question
























  • Reference note: This question, including the terminology, was problem B1 on the 1996 Putnam.
    – jmerry
    Dec 8 at 7:51














4












4








4


1





Define a selfish set to be a set which has its own cardinality (number of elements) as an element. Find, with proof, the number of subsets of ${1, 2, ldots, n}$ which are minimal selfish sets, that is, selfish sets none of whose proper subsets is selfish.



My Attempt:
Assume $textbf{A}$ to be a selfish set. If the cardinality of $textbf{A}$ is $c$, then can $textbf{A}$ contain $1,2,3....c-1$. Definitely answer is no. because if it contains $k<c$ then deleting $c-k$ elements except $k$ from $textbf{A}$ gives a subset of k elements contradicting the fact that $textbf{A}$ is minimal selfish.
Thus $textbf{A}$ must contain elements greater than or equal to $c$. But how do I find the minimal selfish sets with order $c$?






combinatorics discrete-mathematics







share|cite|improve this question















Define a selfish set to be a set which has its own cardinality (number of elements) as an element. Find, with proof, the number of subsets of ${1, 2, ldots, n}$ which are minimal selfish sets, that is, selfish sets none of whose proper subsets is selfish.



My Attempt:
Assume $textbf{A}$ to be a selfish set. If the cardinality of $textbf{A}$ is $c$, then can $textbf{A}$ contain $1,2,3....c-1$. Definitely answer is no. because if it contains $k<c$ then deleting $c-k$ elements except $k$ from $textbf{A}$ gives a subset of k elements contradicting the fact that $textbf{A}$ is minimal selfish.
Thus $textbf{A}$ must contain elements greater than or equal to $c$. But how do I find the minimal selfish sets with order $c$?






combinatorics discrete-mathematics




combinatorics discrete-mathematics






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edited Dec 8 at 10:20









Mutantoe

562411




562411










asked Dec 8 at 6:03







user408906



















  • Reference note: This question, including the terminology, was problem B1 on the 1996 Putnam.
    – jmerry
    Dec 8 at 7:51


















  • Reference note: This question, including the terminology, was problem B1 on the 1996 Putnam.
    – jmerry
    Dec 8 at 7:51
















Reference note: This question, including the terminology, was problem B1 on the 1996 Putnam.
– jmerry
Dec 8 at 7:51




Reference note: This question, including the terminology, was problem B1 on the 1996 Putnam.
– jmerry
Dec 8 at 7:51










2 Answers
2






active

oldest

votes


















4














Your argument is correct.



Lets see if recursion helps.



Let $[n]$ denote the set ${1,2,ldots,n}$, and let $f_n$ denote the
number of minimal selfish subsets of $[n]$. Then the number of
minimal selfish subsets of $[n]$ not containing $n$ is equal to
$f_{n-1}$. On the other hand, for any minimal selfish subset of $[n]$
containing $n$, by subtracting 1 from each element, and then taking
away the element $n-1$ from the set, we obtain a minimal selfish
subset of $[n-2]$ (since $1$ and $n$ cannot both occur in a selfish
set). Conversely, any minimal selfish subset of $[n-2]$ gives rise to
a minimal selfish subset of $[n]$ containing $n$ by the inverse
procedure. Hence the number of minimal selfish subsets of $[n]$
containing $n$ is $f_{n-2}$. Thus we obtain $f_n=f_{n-1}+f_{n-2}$.
Since $f_1=f_2=1$, we have $f_n=F_n$, where $F_n$ denotes the $n$th
term of the Fibonacci sequence.






share|cite|improve this answer





























    1














    Your logic so far is fine. So what you know is that, since $c$ is in the set, then the other $c-1$ elements must all be at least $c+1$. There are $binom{n-c}{c-1}$ ways to choose them.



    Summing over these gives you the total count. It turns out that this gives you the $n^{th}$ Fibonacci number, which you can prove by induction (hint: use Pascal’s identity).






    share|cite|improve this answer





















      Your Answer





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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      Your argument is correct.



      Lets see if recursion helps.



      Let $[n]$ denote the set ${1,2,ldots,n}$, and let $f_n$ denote the
      number of minimal selfish subsets of $[n]$. Then the number of
      minimal selfish subsets of $[n]$ not containing $n$ is equal to
      $f_{n-1}$. On the other hand, for any minimal selfish subset of $[n]$
      containing $n$, by subtracting 1 from each element, and then taking
      away the element $n-1$ from the set, we obtain a minimal selfish
      subset of $[n-2]$ (since $1$ and $n$ cannot both occur in a selfish
      set). Conversely, any minimal selfish subset of $[n-2]$ gives rise to
      a minimal selfish subset of $[n]$ containing $n$ by the inverse
      procedure. Hence the number of minimal selfish subsets of $[n]$
      containing $n$ is $f_{n-2}$. Thus we obtain $f_n=f_{n-1}+f_{n-2}$.
      Since $f_1=f_2=1$, we have $f_n=F_n$, where $F_n$ denotes the $n$th
      term of the Fibonacci sequence.






      share|cite|improve this answer


























        4














        Your argument is correct.



        Lets see if recursion helps.



        Let $[n]$ denote the set ${1,2,ldots,n}$, and let $f_n$ denote the
        number of minimal selfish subsets of $[n]$. Then the number of
        minimal selfish subsets of $[n]$ not containing $n$ is equal to
        $f_{n-1}$. On the other hand, for any minimal selfish subset of $[n]$
        containing $n$, by subtracting 1 from each element, and then taking
        away the element $n-1$ from the set, we obtain a minimal selfish
        subset of $[n-2]$ (since $1$ and $n$ cannot both occur in a selfish
        set). Conversely, any minimal selfish subset of $[n-2]$ gives rise to
        a minimal selfish subset of $[n]$ containing $n$ by the inverse
        procedure. Hence the number of minimal selfish subsets of $[n]$
        containing $n$ is $f_{n-2}$. Thus we obtain $f_n=f_{n-1}+f_{n-2}$.
        Since $f_1=f_2=1$, we have $f_n=F_n$, where $F_n$ denotes the $n$th
        term of the Fibonacci sequence.






        share|cite|improve this answer
























          4












          4








          4






          Your argument is correct.



          Lets see if recursion helps.



          Let $[n]$ denote the set ${1,2,ldots,n}$, and let $f_n$ denote the
          number of minimal selfish subsets of $[n]$. Then the number of
          minimal selfish subsets of $[n]$ not containing $n$ is equal to
          $f_{n-1}$. On the other hand, for any minimal selfish subset of $[n]$
          containing $n$, by subtracting 1 from each element, and then taking
          away the element $n-1$ from the set, we obtain a minimal selfish
          subset of $[n-2]$ (since $1$ and $n$ cannot both occur in a selfish
          set). Conversely, any minimal selfish subset of $[n-2]$ gives rise to
          a minimal selfish subset of $[n]$ containing $n$ by the inverse
          procedure. Hence the number of minimal selfish subsets of $[n]$
          containing $n$ is $f_{n-2}$. Thus we obtain $f_n=f_{n-1}+f_{n-2}$.
          Since $f_1=f_2=1$, we have $f_n=F_n$, where $F_n$ denotes the $n$th
          term of the Fibonacci sequence.






          share|cite|improve this answer












          Your argument is correct.



          Lets see if recursion helps.



          Let $[n]$ denote the set ${1,2,ldots,n}$, and let $f_n$ denote the
          number of minimal selfish subsets of $[n]$. Then the number of
          minimal selfish subsets of $[n]$ not containing $n$ is equal to
          $f_{n-1}$. On the other hand, for any minimal selfish subset of $[n]$
          containing $n$, by subtracting 1 from each element, and then taking
          away the element $n-1$ from the set, we obtain a minimal selfish
          subset of $[n-2]$ (since $1$ and $n$ cannot both occur in a selfish
          set). Conversely, any minimal selfish subset of $[n-2]$ gives rise to
          a minimal selfish subset of $[n]$ containing $n$ by the inverse
          procedure. Hence the number of minimal selfish subsets of $[n]$
          containing $n$ is $f_{n-2}$. Thus we obtain $f_n=f_{n-1}+f_{n-2}$.
          Since $f_1=f_2=1$, we have $f_n=F_n$, where $F_n$ denotes the $n$th
          term of the Fibonacci sequence.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 at 6:10









          Rakesh Bhatt

          851114




          851114























              1














              Your logic so far is fine. So what you know is that, since $c$ is in the set, then the other $c-1$ elements must all be at least $c+1$. There are $binom{n-c}{c-1}$ ways to choose them.



              Summing over these gives you the total count. It turns out that this gives you the $n^{th}$ Fibonacci number, which you can prove by induction (hint: use Pascal’s identity).






              share|cite|improve this answer


























                1














                Your logic so far is fine. So what you know is that, since $c$ is in the set, then the other $c-1$ elements must all be at least $c+1$. There are $binom{n-c}{c-1}$ ways to choose them.



                Summing over these gives you the total count. It turns out that this gives you the $n^{th}$ Fibonacci number, which you can prove by induction (hint: use Pascal’s identity).






                share|cite|improve this answer
























                  1












                  1








                  1






                  Your logic so far is fine. So what you know is that, since $c$ is in the set, then the other $c-1$ elements must all be at least $c+1$. There are $binom{n-c}{c-1}$ ways to choose them.



                  Summing over these gives you the total count. It turns out that this gives you the $n^{th}$ Fibonacci number, which you can prove by induction (hint: use Pascal’s identity).






                  share|cite|improve this answer












                  Your logic so far is fine. So what you know is that, since $c$ is in the set, then the other $c-1$ elements must all be at least $c+1$. There are $binom{n-c}{c-1}$ ways to choose them.



                  Summing over these gives you the total count. It turns out that this gives you the $n^{th}$ Fibonacci number, which you can prove by induction (hint: use Pascal’s identity).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 8 at 6:08









                  platty

                  3,360320




                  3,360320






























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