Can $sum_{i=0}^{n} {i choose k}$ be simplified such that a general calculator can compute its result?
$begingroup$
Is there any way to simplify $sum_{i=0}^{n} {i choose k}$ in such a way that it is computable on a standard calculator similar to how $sum_{i=0}^{n} {n choose i} = 2^n$. I can only seem to find answers relating to the latter equation.
combinatorics
asked Dec 5 '18 at 20:28
Brendan O'SullivanBrendan O'Sullivan
31
$endgroup$
add a comment |
$begingroup$
Is there any way to simplify $sum_{i=0}^{n} {i choose k}$ in such a way that it is computable on a standard calculator similar to how $sum_{i=0}^{n} {n choose i} = 2^n$. I can only seem to find answers relating to the latter equation.
combinatorics
asked Dec 5 '18 at 20:28
Brendan O'SullivanBrendan O'Sullivan
31
$endgroup$
add a comment |
$begingroup$
Is there any way to simplify $sum_{i=0}^{n} {i choose k}$ in such a way that it is computable on a standard calculator similar to how $sum_{i=0}^{n} {n choose i} = 2^n$. I can only seem to find answers relating to the latter equation.
combinatorics
asked Dec 5 '18 at 20:28
Brendan O'SullivanBrendan O'Sullivan
31
$endgroup$
Is there any way to simplify $sum_{i=0}^{n} {i choose k}$ in such a way that it is computable on a standard calculator similar to how $sum_{i=0}^{n} {n choose i} = 2^n$. I can only seem to find answers relating to the latter equation.
combinatorics
combinatorics
asked Dec 5 '18 at 20:28
Brendan O'SullivanBrendan O'Sullivan
31
asked Dec 5 '18 at 20:28
Brendan O'SullivanBrendan O'Sullivan
31
asked Dec 5 '18 at 20:28
Brendan O'SullivanBrendan O'Sullivan
31
asked Dec 5 '18 at 20:28
Brendan O'SullivanBrendan O'Sullivan
31
asked Dec 5 '18 at 20:28
Brendan O'SullivanBrendan O'Sullivan
31
31
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Certainly. In fact:
$$
sum_{i=0}^{n}binom{i}{k}=binom{n+1}{k+1}.
$$
Why? The right side counts the number of ways to choose $k+1$ numbers out of ${1,2,ldots,n+1}$. The $i$th summand on the left counts the number of ways to choose $k+1$ numbers out of ${1,2,ldots,n+1}$ things such that the highest element chosen is $i+1$ (because you then must choose $k$ more elements out of the $i$ elements smaller than $i+1$).
answered Dec 5 '18 at 20:33
Nick PetersonNick Peterson
26.3k23960
$endgroup$
add a comment |
$begingroup$
Yes. It is well-known that
$$
sum_{i=0}^n binom{i}{k}=binom{n+1}{k+1}
$$
by classifying $k+1$ element subsets of ${0, dotsc, n}$ based on their largest element.
answered Dec 5 '18 at 20:31
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
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%2f3027605%2fcan-sum-i-0n-i-choose-k-be-simplified-such-that-a-general-calculator%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Certainly. In fact:
$$
sum_{i=0}^{n}binom{i}{k}=binom{n+1}{k+1}.
$$
Why? The right side counts the number of ways to choose $k+1$ numbers out of ${1,2,ldots,n+1}$. The $i$th summand on the left counts the number of ways to choose $k+1$ numbers out of ${1,2,ldots,n+1}$ things such that the highest element chosen is $i+1$ (because you then must choose $k$ more elements out of the $i$ elements smaller than $i+1$).
answered Dec 5 '18 at 20:33
Nick PetersonNick Peterson
26.3k23960
$endgroup$
add a comment |
$begingroup$
Certainly. In fact:
$$
sum_{i=0}^{n}binom{i}{k}=binom{n+1}{k+1}.
$$
Why? The right side counts the number of ways to choose $k+1$ numbers out of ${1,2,ldots,n+1}$. The $i$th summand on the left counts the number of ways to choose $k+1$ numbers out of ${1,2,ldots,n+1}$ things such that the highest element chosen is $i+1$ (because you then must choose $k$ more elements out of the $i$ elements smaller than $i+1$).
answered Dec 5 '18 at 20:33
Nick PetersonNick Peterson
26.3k23960
$endgroup$
add a comment |
$begingroup$
Certainly. In fact:
$$
sum_{i=0}^{n}binom{i}{k}=binom{n+1}{k+1}.
$$
Why? The right side counts the number of ways to choose $k+1$ numbers out of ${1,2,ldots,n+1}$. The $i$th summand on the left counts the number of ways to choose $k+1$ numbers out of ${1,2,ldots,n+1}$ things such that the highest element chosen is $i+1$ (because you then must choose $k$ more elements out of the $i$ elements smaller than $i+1$).
answered Dec 5 '18 at 20:33
Nick PetersonNick Peterson
26.3k23960
$endgroup$
Certainly. In fact:
$$
sum_{i=0}^{n}binom{i}{k}=binom{n+1}{k+1}.
$$
Why? The right side counts the number of ways to choose $k+1$ numbers out of ${1,2,ldots,n+1}$. The $i$th summand on the left counts the number of ways to choose $k+1$ numbers out of ${1,2,ldots,n+1}$ things such that the highest element chosen is $i+1$ (because you then must choose $k$ more elements out of the $i$ elements smaller than $i+1$).
answered Dec 5 '18 at 20:33
Nick PetersonNick Peterson
26.3k23960
answered Dec 5 '18 at 20:33
Nick PetersonNick Peterson
26.3k23960
answered Dec 5 '18 at 20:33
Nick PetersonNick Peterson
26.3k23960
answered Dec 5 '18 at 20:33
Nick PetersonNick Peterson
26.3k23960
26.3k23960
add a comment |
add a comment |
$begingroup$
Yes. It is well-known that
$$
sum_{i=0}^n binom{i}{k}=binom{n+1}{k+1}
$$
by classifying $k+1$ element subsets of ${0, dotsc, n}$ based on their largest element.
answered Dec 5 '18 at 20:31
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
add a comment |
$begingroup$
Yes. It is well-known that
$$
sum_{i=0}^n binom{i}{k}=binom{n+1}{k+1}
$$
by classifying $k+1$ element subsets of ${0, dotsc, n}$ based on their largest element.
answered Dec 5 '18 at 20:31
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
add a comment |
$begingroup$
Yes. It is well-known that
$$
sum_{i=0}^n binom{i}{k}=binom{n+1}{k+1}
$$
by classifying $k+1$ element subsets of ${0, dotsc, n}$ based on their largest element.
answered Dec 5 '18 at 20:31
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
Yes. It is well-known that
$$
sum_{i=0}^n binom{i}{k}=binom{n+1}{k+1}
$$
by classifying $k+1$ element subsets of ${0, dotsc, n}$ based on their largest element.
answered Dec 5 '18 at 20:31
Foobaz JohnFoobaz John
21.6k41352
answered Dec 5 '18 at 20:31
Foobaz JohnFoobaz John
21.6k41352
answered Dec 5 '18 at 20:31
Foobaz JohnFoobaz John
21.6k41352
answered Dec 5 '18 at 20:31
Foobaz JohnFoobaz John
21.6k41352
21.6k41352
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%2f3027605%2fcan-sum-i-0n-i-choose-k-be-simplified-such-that-a-general-calculator%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
