Combinatorial proof that the number of even cardinality subsets is equal to the number of odd cardinality...












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Given a set of cardinality $ngeq 1$, the number of subsets of even cardinality is equal to the number of subsets of odd cardinality




I am looking for a combinatorial proof of this statement- I know an algebraic proof is easy, for instance by expanding $(1-1)^n$. If $n$ is odd it is easy to see there is a one-to-one correspondence between even-cardinality subsets an odd-cardinality subsets using the complement. However, I can't think of a combinatorial proof if $n$ is even. Any ideas?










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    $begingroup$



    Given a set of cardinality $ngeq 1$, the number of subsets of even cardinality is equal to the number of subsets of odd cardinality




    I am looking for a combinatorial proof of this statement- I know an algebraic proof is easy, for instance by expanding $(1-1)^n$. If $n$ is odd it is easy to see there is a one-to-one correspondence between even-cardinality subsets an odd-cardinality subsets using the complement. However, I can't think of a combinatorial proof if $n$ is even. Any ideas?










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      0





      $begingroup$



      Given a set of cardinality $ngeq 1$, the number of subsets of even cardinality is equal to the number of subsets of odd cardinality




      I am looking for a combinatorial proof of this statement- I know an algebraic proof is easy, for instance by expanding $(1-1)^n$. If $n$ is odd it is easy to see there is a one-to-one correspondence between even-cardinality subsets an odd-cardinality subsets using the complement. However, I can't think of a combinatorial proof if $n$ is even. Any ideas?










      share|cite|improve this question









      $endgroup$





      Given a set of cardinality $ngeq 1$, the number of subsets of even cardinality is equal to the number of subsets of odd cardinality




      I am looking for a combinatorial proof of this statement- I know an algebraic proof is easy, for instance by expanding $(1-1)^n$. If $n$ is odd it is easy to see there is a one-to-one correspondence between even-cardinality subsets an odd-cardinality subsets using the complement. However, I can't think of a combinatorial proof if $n$ is even. Any ideas?







      combinatorial-proofs






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      asked Dec 19 '18 at 6:39









      Darren OngDarren Ong

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          Just pick your favourite element $a$. Now take a subset $X$ and add $a$ if $anotin X$
          and delete it if $ain X$. One gets a pairing of the subsets, and in each pair
          one subset is even, the other odd.






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            use the case for odd $n$, and consider adding a single additional element to the set. the old subsets are still subsets, together with new ones which contain the additional element.






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              2 Answers
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              2 Answers
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              $begingroup$

              Just pick your favourite element $a$. Now take a subset $X$ and add $a$ if $anotin X$
              and delete it if $ain X$. One gets a pairing of the subsets, and in each pair
              one subset is even, the other odd.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Just pick your favourite element $a$. Now take a subset $X$ and add $a$ if $anotin X$
                and delete it if $ain X$. One gets a pairing of the subsets, and in each pair
                one subset is even, the other odd.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Just pick your favourite element $a$. Now take a subset $X$ and add $a$ if $anotin X$
                  and delete it if $ain X$. One gets a pairing of the subsets, and in each pair
                  one subset is even, the other odd.






                  share|cite|improve this answer









                  $endgroup$



                  Just pick your favourite element $a$. Now take a subset $X$ and add $a$ if $anotin X$
                  and delete it if $ain X$. One gets a pairing of the subsets, and in each pair
                  one subset is even, the other odd.







                  share|cite|improve this answer












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                  answered Dec 19 '18 at 6:49









                  Lord Shark the UnknownLord Shark the Unknown

                  107k1162135




                  107k1162135























                      0












                      $begingroup$

                      use the case for odd $n$, and consider adding a single additional element to the set. the old subsets are still subsets, together with new ones which contain the additional element.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        use the case for odd $n$, and consider adding a single additional element to the set. the old subsets are still subsets, together with new ones which contain the additional element.






                        share|cite|improve this answer









                        $endgroup$
















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                          $begingroup$

                          use the case for odd $n$, and consider adding a single additional element to the set. the old subsets are still subsets, together with new ones which contain the additional element.






                          share|cite|improve this answer









                          $endgroup$



                          use the case for odd $n$, and consider adding a single additional element to the set. the old subsets are still subsets, together with new ones which contain the additional element.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 19 '18 at 6:48









                          David HoldenDavid Holden

                          14.9k21225




                          14.9k21225






























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