Combinatorics Proof?
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How would I prove
$$sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} = binom{2a}{a}binom{2a}{a}$$
I am familiar with the identity $$binom{2n}{n}=sum_{k=0}^nbinom{n}{k}binom{n}{n-k}=binom{n}{k}^2$$ Would I be able to use this?
combinatorics proof-verification
add a comment |
up vote
3
down vote
favorite
How would I prove
$$sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} = binom{2a}{a}binom{2a}{a}$$
I am familiar with the identity $$binom{2n}{n}=sum_{k=0}^nbinom{n}{k}binom{n}{n-k}=binom{n}{k}^2$$ Would I be able to use this?
combinatorics proof-verification
I would say yes. Very much so.
– Somos
Dec 11 at 17:04
and in the last identity, you lost the Sum
– G Cab
Dec 11 at 17:45
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
How would I prove
$$sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} = binom{2a}{a}binom{2a}{a}$$
I am familiar with the identity $$binom{2n}{n}=sum_{k=0}^nbinom{n}{k}binom{n}{n-k}=binom{n}{k}^2$$ Would I be able to use this?
combinatorics proof-verification
How would I prove
$$sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} = binom{2a}{a}binom{2a}{a}$$
I am familiar with the identity $$binom{2n}{n}=sum_{k=0}^nbinom{n}{k}binom{n}{n-k}=binom{n}{k}^2$$ Would I be able to use this?
combinatorics proof-verification
combinatorics proof-verification
edited Dec 11 at 22:18
asked Dec 11 at 16:52
user497933
I would say yes. Very much so.
– Somos
Dec 11 at 17:04
and in the last identity, you lost the Sum
– G Cab
Dec 11 at 17:45
add a comment |
I would say yes. Very much so.
– Somos
Dec 11 at 17:04
and in the last identity, you lost the Sum
– G Cab
Dec 11 at 17:45
I would say yes. Very much so.
– Somos
Dec 11 at 17:04
I would say yes. Very much so.
– Somos
Dec 11 at 17:04
and in the last identity, you lost the Sum
– G Cab
Dec 11 at 17:45
and in the last identity, you lost the Sum
– G Cab
Dec 11 at 17:45
add a comment |
2 Answers
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up vote
6
down vote
This is not purely bijective proof. The identity that you said you knew of it can be proved bijectively. We will use that fact to prove the new identity.
First rewrite
begin{align}
frac{(2a)!}{b!b!(a-b)!(a-b)!} &= frac{(2a)!}{a!a!}frac{a!a!}{b!(b-a)!b!(b-a)!}\
&={2a choose a}
{achoose b}{a choose b-a}end{align}
It follows that
begin{align}
sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} &={2a choose a}sum_{b=0}^a {achoose b}{achoose a-b}\
&={2achoose a}{2achoose a}quad text{(by your identity)}
end{align}
answered Dec 11 at 17:21
user9077
1,254612
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up vote
2
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The summand $frac{(2a)!}{(b!)^2((a-b)!)^2}$ counts the number of ways to color the elements of a set $S$ of size $2a$ in four colors, so that there are $b$ elements in the first two colors and $a-b$ elements in the last two colors.
The right hand side counts the number of ways to choose two subsets $A$ and $B$ of $S$ of size $a$. This induces a coloring in the following way:
$xin A, xin Bimplies x$ is red.
$xin A, xnotin Bimplies x$ is blue.
$xnotin A, xin Bimplies x$ is green.
$xnotin A, xnotin Bimplies x$ is black.
Combining these two pieces, you can make a combinatorial proof.
answered Dec 11 at 17:57
Mike Earnest
19.9k11950
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
This is not purely bijective proof. The identity that you said you knew of it can be proved bijectively. We will use that fact to prove the new identity.
First rewrite
begin{align}
frac{(2a)!}{b!b!(a-b)!(a-b)!} &= frac{(2a)!}{a!a!}frac{a!a!}{b!(b-a)!b!(b-a)!}\
&={2a choose a}
{achoose b}{a choose b-a}end{align}
It follows that
begin{align}
sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} &={2a choose a}sum_{b=0}^a {achoose b}{achoose a-b}\
&={2achoose a}{2achoose a}quad text{(by your identity)}
end{align}
answered Dec 11 at 17:21
user9077
1,254612
add a comment |
up vote
6
down vote
This is not purely bijective proof. The identity that you said you knew of it can be proved bijectively. We will use that fact to prove the new identity.
First rewrite
begin{align}
frac{(2a)!}{b!b!(a-b)!(a-b)!} &= frac{(2a)!}{a!a!}frac{a!a!}{b!(b-a)!b!(b-a)!}\
&={2a choose a}
{achoose b}{a choose b-a}end{align}
It follows that
begin{align}
sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} &={2a choose a}sum_{b=0}^a {achoose b}{achoose a-b}\
&={2achoose a}{2achoose a}quad text{(by your identity)}
end{align}
answered Dec 11 at 17:21
user9077
1,254612
add a comment |
up vote
6
down vote
up vote
6
down vote
This is not purely bijective proof. The identity that you said you knew of it can be proved bijectively. We will use that fact to prove the new identity.
First rewrite
begin{align}
frac{(2a)!}{b!b!(a-b)!(a-b)!} &= frac{(2a)!}{a!a!}frac{a!a!}{b!(b-a)!b!(b-a)!}\
&={2a choose a}
{achoose b}{a choose b-a}end{align}
It follows that
begin{align}
sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} &={2a choose a}sum_{b=0}^a {achoose b}{achoose a-b}\
&={2achoose a}{2achoose a}quad text{(by your identity)}
end{align}
answered Dec 11 at 17:21
user9077
1,254612
This is not purely bijective proof. The identity that you said you knew of it can be proved bijectively. We will use that fact to prove the new identity.
First rewrite
begin{align}
frac{(2a)!}{b!b!(a-b)!(a-b)!} &= frac{(2a)!}{a!a!}frac{a!a!}{b!(b-a)!b!(b-a)!}\
&={2a choose a}
{achoose b}{a choose b-a}end{align}
It follows that
begin{align}
sum_{b=0}^a frac{(2a)!}{b!b!(a-b)!(a-b)!} &={2a choose a}sum_{b=0}^a {achoose b}{achoose a-b}\
&={2achoose a}{2achoose a}quad text{(by your identity)}
end{align}
answered Dec 11 at 17:21
user9077
1,254612
answered Dec 11 at 17:21
user9077
1,254612
answered Dec 11 at 17:21
user9077
1,254612
answered Dec 11 at 17:21
user9077
1,254612
1,254612
add a comment |
add a comment |
up vote
2
down vote
The summand $frac{(2a)!}{(b!)^2((a-b)!)^2}$ counts the number of ways to color the elements of a set $S$ of size $2a$ in four colors, so that there are $b$ elements in the first two colors and $a-b$ elements in the last two colors.
The right hand side counts the number of ways to choose two subsets $A$ and $B$ of $S$ of size $a$. This induces a coloring in the following way:
$xin A, xin Bimplies x$ is red.
$xin A, xnotin Bimplies x$ is blue.
$xnotin A, xin Bimplies x$ is green.
$xnotin A, xnotin Bimplies x$ is black.
Combining these two pieces, you can make a combinatorial proof.
answered Dec 11 at 17:57
Mike Earnest
19.9k11950
add a comment |
up vote
2
down vote
The summand $frac{(2a)!}{(b!)^2((a-b)!)^2}$ counts the number of ways to color the elements of a set $S$ of size $2a$ in four colors, so that there are $b$ elements in the first two colors and $a-b$ elements in the last two colors.
The right hand side counts the number of ways to choose two subsets $A$ and $B$ of $S$ of size $a$. This induces a coloring in the following way:
$xin A, xin Bimplies x$ is red.
$xin A, xnotin Bimplies x$ is blue.
$xnotin A, xin Bimplies x$ is green.
$xnotin A, xnotin Bimplies x$ is black.
Combining these two pieces, you can make a combinatorial proof.
answered Dec 11 at 17:57
Mike Earnest
19.9k11950
add a comment |
up vote
2
down vote
up vote
2
down vote
The summand $frac{(2a)!}{(b!)^2((a-b)!)^2}$ counts the number of ways to color the elements of a set $S$ of size $2a$ in four colors, so that there are $b$ elements in the first two colors and $a-b$ elements in the last two colors.
The right hand side counts the number of ways to choose two subsets $A$ and $B$ of $S$ of size $a$. This induces a coloring in the following way:
$xin A, xin Bimplies x$ is red.
$xin A, xnotin Bimplies x$ is blue.
$xnotin A, xin Bimplies x$ is green.
$xnotin A, xnotin Bimplies x$ is black.
Combining these two pieces, you can make a combinatorial proof.
answered Dec 11 at 17:57
Mike Earnest
19.9k11950
The summand $frac{(2a)!}{(b!)^2((a-b)!)^2}$ counts the number of ways to color the elements of a set $S$ of size $2a$ in four colors, so that there are $b$ elements in the first two colors and $a-b$ elements in the last two colors.
The right hand side counts the number of ways to choose two subsets $A$ and $B$ of $S$ of size $a$. This induces a coloring in the following way:
$xin A, xin Bimplies x$ is red.
$xin A, xnotin Bimplies x$ is blue.
$xnotin A, xin Bimplies x$ is green.
$xnotin A, xnotin Bimplies x$ is black.
Combining these two pieces, you can make a combinatorial proof.
answered Dec 11 at 17:57
Mike Earnest
19.9k11950
answered Dec 11 at 17:57
Mike Earnest
19.9k11950
answered Dec 11 at 17:57
Mike Earnest
19.9k11950
answered Dec 11 at 17:57
Mike Earnest
19.9k11950
19.9k11950
add a comment |
add a comment |
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I would say yes. Very much so.
– Somos
Dec 11 at 17:04
and in the last identity, you lost the Sum
– G Cab
Dec 11 at 17:45