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?
combinatorial-proofs
asked Dec 19 '18 at 6:39
Darren OngDarren Ong
27819
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add a comment |
$begingroup$
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?
combinatorial-proofs
asked Dec 19 '18 at 6:39
Darren OngDarren Ong
27819
$endgroup$
add a comment |
$begingroup$
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?
combinatorial-proofs
asked Dec 19 '18 at 6:39
Darren OngDarren Ong
27819
$endgroup$
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
combinatorial-proofs
asked Dec 19 '18 at 6:39
Darren OngDarren Ong
27819
asked Dec 19 '18 at 6:39
Darren OngDarren Ong
27819
asked Dec 19 '18 at 6:39
Darren OngDarren Ong
27819
asked Dec 19 '18 at 6:39
Darren OngDarren Ong
27819
asked Dec 19 '18 at 6:39
Darren OngDarren Ong
27819
27819
add a comment |
add a comment |
2 Answers
2
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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.
answered Dec 19 '18 at 6:49
Lord Shark the UnknownLord Shark the Unknown
103k1160132
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add a comment |
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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.
answered Dec 19 '18 at 6:48
David HoldenDavid Holden
14.7k21224
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add a comment |
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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.
answered Dec 19 '18 at 6:49
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 6:49
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 6:49
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
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.
answered Dec 19 '18 at 6:49
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Dec 19 '18 at 6:49
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Dec 19 '18 at 6:49
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Dec 19 '18 at 6:49
Lord Shark the UnknownLord Shark the Unknown
103k1160132
103k1160132
add a comment |
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 6:48
David HoldenDavid Holden
14.7k21224
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 6:48
David HoldenDavid Holden
14.7k21224
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 6:48
David HoldenDavid Holden
14.7k21224
$endgroup$
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.
answered Dec 19 '18 at 6:48
David HoldenDavid Holden
14.7k21224
answered Dec 19 '18 at 6:48
David HoldenDavid Holden
14.7k21224
answered Dec 19 '18 at 6:48
David HoldenDavid Holden
14.7k21224
answered Dec 19 '18 at 6:48
David HoldenDavid Holden
14.7k21224
14.7k21224
add a comment |
add a comment |
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