How many ways are there to choose k numbers from first n natural numbers such that any two numbers are at...
$begingroup$
How many ways are there to choose $k$ distinct numbers from $n$ consecutive natural numbers such that any two numbers are at most differ by $d$?
The solution of the similar problem is done when difference is at least $d$.But what about when the difference is at most $d$?
combinatorics
$endgroup$
add a comment |
$begingroup$
How many ways are there to choose $k$ distinct numbers from $n$ consecutive natural numbers such that any two numbers are at most differ by $d$?
The solution of the similar problem is done when difference is at least $d$.But what about when the difference is at most $d$?
combinatorics
$endgroup$
add a comment |
$begingroup$
How many ways are there to choose $k$ distinct numbers from $n$ consecutive natural numbers such that any two numbers are at most differ by $d$?
The solution of the similar problem is done when difference is at least $d$.But what about when the difference is at most $d$?
combinatorics
$endgroup$
How many ways are there to choose $k$ distinct numbers from $n$ consecutive natural numbers such that any two numbers are at most differ by $d$?
The solution of the similar problem is done when difference is at least $d$.But what about when the difference is at most $d$?
combinatorics
combinatorics
asked Jan 7 at 12:35
Supriyo HalderSupriyo Halder
633113
633113
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assume WLOG that the numbers are $1, 2, ..., n$ and assume $k-1leq d<n$.
Let's focus on numbers $1, 2, ..., d+1$. Obviously, the difference between any two numbers is at most $d$. We can choose $k$ numbers in $binom{d+1}{k}$ ways.
Now focus on numbers $2, 3, ..., d+2$. Since we have already considered all $k-subsets$ on the first $d$ numbers, we have to include $d+2$ in our new choices, hence, leading to $binom{d}{k-1}$ ways.
Focus on $3, 4, ..., d+3$. Same logic as above.
Repeat this $n-d$ times and get a total of
$$underbrace{binom{d+1}{k} + binom{d}{k-1} + ... + binom{d}{k-1}}_text{n-d terms}=binom{d+1}{k}+(n-d-1)binom{d}{k-1}$$.
$endgroup$
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%2f3064963%2fhow-many-ways-are-there-to-choose-k-numbers-from-first-n-natural-numbers-such-th%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
$begingroup$
Assume WLOG that the numbers are $1, 2, ..., n$ and assume $k-1leq d<n$.
Let's focus on numbers $1, 2, ..., d+1$. Obviously, the difference between any two numbers is at most $d$. We can choose $k$ numbers in $binom{d+1}{k}$ ways.
Now focus on numbers $2, 3, ..., d+2$. Since we have already considered all $k-subsets$ on the first $d$ numbers, we have to include $d+2$ in our new choices, hence, leading to $binom{d}{k-1}$ ways.
Focus on $3, 4, ..., d+3$. Same logic as above.
Repeat this $n-d$ times and get a total of
$$underbrace{binom{d+1}{k} + binom{d}{k-1} + ... + binom{d}{k-1}}_text{n-d terms}=binom{d+1}{k}+(n-d-1)binom{d}{k-1}$$.
$endgroup$
add a comment |
$begingroup$
Assume WLOG that the numbers are $1, 2, ..., n$ and assume $k-1leq d<n$.
Let's focus on numbers $1, 2, ..., d+1$. Obviously, the difference between any two numbers is at most $d$. We can choose $k$ numbers in $binom{d+1}{k}$ ways.
Now focus on numbers $2, 3, ..., d+2$. Since we have already considered all $k-subsets$ on the first $d$ numbers, we have to include $d+2$ in our new choices, hence, leading to $binom{d}{k-1}$ ways.
Focus on $3, 4, ..., d+3$. Same logic as above.
Repeat this $n-d$ times and get a total of
$$underbrace{binom{d+1}{k} + binom{d}{k-1} + ... + binom{d}{k-1}}_text{n-d terms}=binom{d+1}{k}+(n-d-1)binom{d}{k-1}$$.
$endgroup$
add a comment |
$begingroup$
Assume WLOG that the numbers are $1, 2, ..., n$ and assume $k-1leq d<n$.
Let's focus on numbers $1, 2, ..., d+1$. Obviously, the difference between any two numbers is at most $d$. We can choose $k$ numbers in $binom{d+1}{k}$ ways.
Now focus on numbers $2, 3, ..., d+2$. Since we have already considered all $k-subsets$ on the first $d$ numbers, we have to include $d+2$ in our new choices, hence, leading to $binom{d}{k-1}$ ways.
Focus on $3, 4, ..., d+3$. Same logic as above.
Repeat this $n-d$ times and get a total of
$$underbrace{binom{d+1}{k} + binom{d}{k-1} + ... + binom{d}{k-1}}_text{n-d terms}=binom{d+1}{k}+(n-d-1)binom{d}{k-1}$$.
$endgroup$
Assume WLOG that the numbers are $1, 2, ..., n$ and assume $k-1leq d<n$.
Let's focus on numbers $1, 2, ..., d+1$. Obviously, the difference between any two numbers is at most $d$. We can choose $k$ numbers in $binom{d+1}{k}$ ways.
Now focus on numbers $2, 3, ..., d+2$. Since we have already considered all $k-subsets$ on the first $d$ numbers, we have to include $d+2$ in our new choices, hence, leading to $binom{d}{k-1}$ ways.
Focus on $3, 4, ..., d+3$. Same logic as above.
Repeat this $n-d$ times and get a total of
$$underbrace{binom{d+1}{k} + binom{d}{k-1} + ... + binom{d}{k-1}}_text{n-d terms}=binom{d+1}{k}+(n-d-1)binom{d}{k-1}$$.
answered Jan 7 at 13:10
EuxhenHEuxhenH
484210
484210
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%2f3064963%2fhow-many-ways-are-there-to-choose-k-numbers-from-first-n-natural-numbers-such-th%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