Mixing
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
48.7k870156
asked Jan 20 at 20:28
ZacharyZachary
1799
$endgroup$
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
48.7k870156
asked Jan 20 at 20:28
ZacharyZachary
1799
$endgroup$
$begingroup$
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
48.7k870156
asked Jan 20 at 20:28
ZacharyZachary
1799
$endgroup$
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
48.7k870156
asked Jan 20 at 20:28
ZacharyZachary
1799
edited Jan 22 at 19:09
Blue
48.7k870156
asked Jan 20 at 20:28
ZacharyZachary
1799
edited Jan 22 at 19:09
Blue
48.7k870156
edited Jan 22 at 19:09
Blue
48.7k870156
edited Jan 22 at 19:09
Blue
48.7k870156
48.7k870156
asked Jan 20 at 20:28
ZacharyZachary
1799
asked Jan 20 at 20:28
ZacharyZachary
1799
asked Jan 20 at 20:28
ZacharyZachary
1799
1799
$begingroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3081098%2fcombinatoric-proof-that-sum-k-0m-binommk-binomnkm-sum-k-0m%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 |
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3081098%2fcombinatoric-proof-that-sum-k-0m-binommk-binomnkm-sum-k-0m%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
What have you tried?
$endgroup$
– Phicar
Jan 22 at 2:52