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
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add a comment |
$begingroup$
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
$endgroup$
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I would say yes. Very much so.
$endgroup$
– Somos
Dec 11 '18 at 17:04
$begingroup$
and in the last identity, you lost the Sum
$endgroup$
– G Cab
Dec 11 '18 at 17:45
add a comment |
$begingroup$
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
$endgroup$
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 '18 at 22:18
asked Dec 11 '18 at 16:52
user497933
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I would say yes. Very much so.
$endgroup$
– Somos
Dec 11 '18 at 17:04
$begingroup$
and in the last identity, you lost the Sum
$endgroup$
– G Cab
Dec 11 '18 at 17:45
add a comment |
$begingroup$
I would say yes. Very much so.
$endgroup$
– Somos
Dec 11 '18 at 17:04
$begingroup$
and in the last identity, you lost the Sum
$endgroup$
– G Cab
Dec 11 '18 at 17:45
$begingroup$
I would say yes. Very much so.
$endgroup$
– Somos
Dec 11 '18 at 17:04
$begingroup$
I would say yes. Very much so.
$endgroup$
– Somos
Dec 11 '18 at 17:04
$begingroup$
and in the last identity, you lost the Sum
$endgroup$
– G Cab
Dec 11 '18 at 17:45
$begingroup$
and in the last identity, you lost the Sum
$endgroup$
– G Cab
Dec 11 '18 at 17:45
add a comment |
2 Answers
2
active
oldest
votes
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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}
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add a comment |
$begingroup$
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.
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add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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}
$endgroup$
add a comment |
$begingroup$
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}
$endgroup$
add a comment |
$begingroup$
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}
$endgroup$
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 '18 at 17:21
user9077user9077
1,239612
1,239612
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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 '18 at 17:57
Mike EarnestMike Earnest
22.6k12051
22.6k12051
add a comment |
add a comment |
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$begingroup$
I would say yes. Very much so.
$endgroup$
– Somos
Dec 11 '18 at 17:04
$begingroup$
and in the last identity, you lost the Sum
$endgroup$
– G Cab
Dec 11 '18 at 17:45