Combinatoric proof that $sum_{k=0}^m binom{m}{k}binom{n+k}{m} = sum_{k=0}^m binom{n}{k}binom{m}{m-k}2^k$
$begingroup$
Proof that:
$$sum_{k=0}^m binom{m}{k}binom{n+k}{m} = sum_{k=0}^m binom{n}{k}binom{m}{m-k}2^k$$
for $k,m,n in mathbb{N}$.
Can someone help me understand this equality by giving a combinatoric argument?
I believe that the RHS means "all possible ways to first select $k$ elements from $m$ elements and thereafter, selecting $m$ elements out of $n+k$ elements". Same for the LHS. The factor $2^k$ probably has to do something with having to pick between two possible options for $k$ elements.
I can't really find a good scenario to prove the equation. I tried with a group of $m$ kids, with $n$ boys and $k$ girls. But that did not really work out.
Thanks for helping!
combinatorics
edited Jan 22 at 19:09
Blue
48.6k870156
asked Jan 20 at 20:28
ZacharyZachary
1749
$endgroup$
add a comment |
$begingroup$
Proof that:
$$sum_{k=0}^m binom{m}{k}binom{n+k}{m} = sum_{k=0}^m binom{n}{k}binom{m}{m-k}2^k$$
for $k,m,n in mathbb{N}$.
Can someone help me understand this equality by giving a combinatoric argument?
I believe that the RHS means "all possible ways to first select $k$ elements from $m$ elements and thereafter, selecting $m$ elements out of $n+k$ elements". Same for the LHS. The factor $2^k$ probably has to do something with having to pick between two possible options for $k$ elements.
I can't really find a good scenario to prove the equation. I tried with a group of $m$ kids, with $n$ boys and $k$ girls. But that did not really work out.
Thanks for helping!
combinatorics
edited Jan 22 at 19:09
Blue
48.6k870156
asked Jan 20 at 20:28
ZacharyZachary
1749
$endgroup$
$begingroup$
What have you tried?
$endgroup$
– Phicar
Jan 22 at 2:52
add a comment |
$begingroup$
Proof that:
$$sum_{k=0}^m binom{m}{k}binom{n+k}{m} = sum_{k=0}^m binom{n}{k}binom{m}{m-k}2^k$$
for $k,m,n in mathbb{N}$.
Can someone help me understand this equality by giving a combinatoric argument?
I believe that the RHS means "all possible ways to first select $k$ elements from $m$ elements and thereafter, selecting $m$ elements out of $n+k$ elements". Same for the LHS. The factor $2^k$ probably has to do something with having to pick between two possible options for $k$ elements.
I can't really find a good scenario to prove the equation. I tried with a group of $m$ kids, with $n$ boys and $k$ girls. But that did not really work out.
Thanks for helping!
combinatorics
edited Jan 22 at 19:09
Blue
48.6k870156
asked Jan 20 at 20:28
ZacharyZachary
1749
$endgroup$
Proof that:
$$sum_{k=0}^m binom{m}{k}binom{n+k}{m} = sum_{k=0}^m binom{n}{k}binom{m}{m-k}2^k$$
for $k,m,n in mathbb{N}$.
Can someone help me understand this equality by giving a combinatoric argument?
I believe that the RHS means "all possible ways to first select $k$ elements from $m$ elements and thereafter, selecting $m$ elements out of $n+k$ elements". Same for the LHS. The factor $2^k$ probably has to do something with having to pick between two possible options for $k$ elements.
I can't really find a good scenario to prove the equation. I tried with a group of $m$ kids, with $n$ boys and $k$ girls. But that did not really work out.
Thanks for helping!
combinatorics
combinatorics
edited Jan 22 at 19:09
Blue
48.6k870156
asked Jan 20 at 20:28
ZacharyZachary
1749
edited Jan 22 at 19:09
Blue
48.6k870156
asked Jan 20 at 20:28
ZacharyZachary
1749
edited Jan 22 at 19:09
Blue
48.6k870156
edited Jan 22 at 19:09
Blue
48.6k870156
edited Jan 22 at 19:09
Blue
48.6k870156
48.6k870156
asked Jan 20 at 20:28
ZacharyZachary
1749
asked Jan 20 at 20:28
ZacharyZachary
1749
asked Jan 20 at 20:28
ZacharyZachary
1749
1749
$begingroup$
What have you tried?
$endgroup$
– Phicar
Jan 22 at 2:52
add a comment |
$begingroup$
What have you tried?
$endgroup$
– Phicar
Jan 22 at 2:52
$begingroup$
What have you tried?
$endgroup$
– Phicar
Jan 22 at 2:52
$begingroup$
What have you tried?
$endgroup$
– Phicar
Jan 22 at 2:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Well suppose you have $m$ boys and $n$ girls and you want to pick a team of $m$ people. Assume, further, that the way to select them is as follows, you first pick $k$ man in a preliminary round $binom{m}{k}$ and in a secondary round out of the chosen man and all the woman you pick your team of $m$ i.e., $binom{n+k}{m}.$ Now, assume you pick directly the man and the women of the team, so they have to add up to $m$ so $binom{n}{k}binom{m}{m-k}$ but you have to pay the men who pass the preliminary round in the LHS, and so you can choose any subset of the man who were not selected ($k$ of them) in $2^k$ ways.
answered Jan 22 at 19:02
PhicarPhicar
2,6401915
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Well suppose you have $m$ boys and $n$ girls and you want to pick a team of $m$ people. Assume, further, that the way to select them is as follows, you first pick $k$ man in a preliminary round $binom{m}{k}$ and in a secondary round out of the chosen man and all the woman you pick your team of $m$ i.e., $binom{n+k}{m}.$ Now, assume you pick directly the man and the women of the team, so they have to add up to $m$ so $binom{n}{k}binom{m}{m-k}$ but you have to pay the men who pass the preliminary round in the LHS, and so you can choose any subset of the man who were not selected ($k$ of them) in $2^k$ ways.
answered Jan 22 at 19:02
PhicarPhicar
2,6401915
$endgroup$
add a comment |
$begingroup$
Well suppose you have $m$ boys and $n$ girls and you want to pick a team of $m$ people. Assume, further, that the way to select them is as follows, you first pick $k$ man in a preliminary round $binom{m}{k}$ and in a secondary round out of the chosen man and all the woman you pick your team of $m$ i.e., $binom{n+k}{m}.$ Now, assume you pick directly the man and the women of the team, so they have to add up to $m$ so $binom{n}{k}binom{m}{m-k}$ but you have to pay the men who pass the preliminary round in the LHS, and so you can choose any subset of the man who were not selected ($k$ of them) in $2^k$ ways.
answered Jan 22 at 19:02
PhicarPhicar
2,6401915
$endgroup$
add a comment |
$begingroup$
Well suppose you have $m$ boys and $n$ girls and you want to pick a team of $m$ people. Assume, further, that the way to select them is as follows, you first pick $k$ man in a preliminary round $binom{m}{k}$ and in a secondary round out of the chosen man and all the woman you pick your team of $m$ i.e., $binom{n+k}{m}.$ Now, assume you pick directly the man and the women of the team, so they have to add up to $m$ so $binom{n}{k}binom{m}{m-k}$ but you have to pay the men who pass the preliminary round in the LHS, and so you can choose any subset of the man who were not selected ($k$ of them) in $2^k$ ways.
answered Jan 22 at 19:02
PhicarPhicar
2,6401915
$endgroup$
Well suppose you have $m$ boys and $n$ girls and you want to pick a team of $m$ people. Assume, further, that the way to select them is as follows, you first pick $k$ man in a preliminary round $binom{m}{k}$ and in a secondary round out of the chosen man and all the woman you pick your team of $m$ i.e., $binom{n+k}{m}.$ Now, assume you pick directly the man and the women of the team, so they have to add up to $m$ so $binom{n}{k}binom{m}{m-k}$ but you have to pay the men who pass the preliminary round in the LHS, and so you can choose any subset of the man who were not selected ($k$ of them) in $2^k$ ways.
answered Jan 22 at 19:02
PhicarPhicar
2,6401915
answered Jan 22 at 19:02
PhicarPhicar
2,6401915
answered Jan 22 at 19:02
PhicarPhicar
2,6401915
answered Jan 22 at 19:02
PhicarPhicar
2,6401915
2,6401915
add a comment |
add a comment |
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$begingroup$
What have you tried?
$endgroup$
– Phicar
Jan 22 at 2:52