In how many ways can I write $0$ as a sum of $n; 0s, 1s ;text{and}; -1s?$












3












$begingroup$


In how many ways can I write $0$ as a sum of $n; 0s, 1s ;text{or}; -1s?$ (Taking the order into account).
I suspect there is no closed formula to express the result, but I'd like someone to confirm it, or deny it.



Edit:

e.g., if $n=3$
$$begin{aligned}0&=;;;0+0+0,\ 0&=;;;1-1+0,\ 0&=;;;1+0-1,\ 0&=;;;0+1-1,\ 0&=-1+1+0,\ 0&=-1+0+1,\ 0&=;;;0-1+1.end{aligned}$$
So for $n=3;$ there are $7$ ways.






combinatorics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Set up a recurrence perhaps.
    $endgroup$
    – paw88789
    Dec 29 '18 at 16:35












  • $begingroup$
    do you mean n each or a total of n?
    $endgroup$
    – player100
    Dec 29 '18 at 16:45










  • $begingroup$
    Sorry I meant that in total, the number of 0's, 1's and -1's is n
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 16:59










  • $begingroup$
    @MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
    $endgroup$
    – timtfj
    Dec 29 '18 at 17:11










  • $begingroup$
    I'd start here: for each $n$, we can choose how many times $0$ occurs ,and how many times $1$ occurs. $-1$ then has to match $1$. So if $0$ is used $r$ times and $1$ is used $s$ times, $r=n-2s$.Then for each choice of $r$ and $s$ you've got a defined permutation problem, with $r, s$ and $s$ items..
    $endgroup$
    – timtfj
    Dec 29 '18 at 17:25


















3












$begingroup$


In how many ways can I write $0$ as a sum of $n; 0s, 1s ;text{or}; -1s?$ (Taking the order into account).
I suspect there is no closed formula to express the result, but I'd like someone to confirm it, or deny it.



Edit:

e.g., if $n=3$
$$begin{aligned}0&=;;;0+0+0,\ 0&=;;;1-1+0,\ 0&=;;;1+0-1,\ 0&=;;;0+1-1,\ 0&=-1+1+0,\ 0&=-1+0+1,\ 0&=;;;0-1+1.end{aligned}$$
So for $n=3;$ there are $7$ ways.






combinatorics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Set up a recurrence perhaps.
    $endgroup$
    – paw88789
    Dec 29 '18 at 16:35












  • $begingroup$
    do you mean n each or a total of n?
    $endgroup$
    – player100
    Dec 29 '18 at 16:45










  • $begingroup$
    Sorry I meant that in total, the number of 0's, 1's and -1's is n
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 16:59










  • $begingroup$
    @MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
    $endgroup$
    – timtfj
    Dec 29 '18 at 17:11










  • $begingroup$
    I'd start here: for each $n$, we can choose how many times $0$ occurs ,and how many times $1$ occurs. $-1$ then has to match $1$. So if $0$ is used $r$ times and $1$ is used $s$ times, $r=n-2s$.Then for each choice of $r$ and $s$ you've got a defined permutation problem, with $r, s$ and $s$ items..
    $endgroup$
    – timtfj
    Dec 29 '18 at 17:25
















3












3








3


2



$begingroup$


In how many ways can I write $0$ as a sum of $n; 0s, 1s ;text{or}; -1s?$ (Taking the order into account).
I suspect there is no closed formula to express the result, but I'd like someone to confirm it, or deny it.



Edit:

e.g., if $n=3$
$$begin{aligned}0&=;;;0+0+0,\ 0&=;;;1-1+0,\ 0&=;;;1+0-1,\ 0&=;;;0+1-1,\ 0&=-1+1+0,\ 0&=-1+0+1,\ 0&=;;;0-1+1.end{aligned}$$
So for $n=3;$ there are $7$ ways.






combinatorics







share|cite|improve this question











$endgroup$




In how many ways can I write $0$ as a sum of $n; 0s, 1s ;text{or}; -1s?$ (Taking the order into account).
I suspect there is no closed formula to express the result, but I'd like someone to confirm it, or deny it.



Edit:

e.g., if $n=3$
$$begin{aligned}0&=;;;0+0+0,\ 0&=;;;1-1+0,\ 0&=;;;1+0-1,\ 0&=;;;0+1-1,\ 0&=-1+1+0,\ 0&=-1+0+1,\ 0&=;;;0-1+1.end{aligned}$$
So for $n=3;$ there are $7$ ways.






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 22:01









user376343

3,2982825




3,2982825










asked Dec 29 '18 at 16:31









Lucio TanziniLucio Tanzini

18414




18414








  • 1




    $begingroup$
    Set up a recurrence perhaps.
    $endgroup$
    – paw88789
    Dec 29 '18 at 16:35












  • $begingroup$
    do you mean n each or a total of n?
    $endgroup$
    – player100
    Dec 29 '18 at 16:45










  • $begingroup$
    Sorry I meant that in total, the number of 0's, 1's and -1's is n
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 16:59










  • $begingroup$
    @MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
    $endgroup$
    – timtfj
    Dec 29 '18 at 17:11










  • $begingroup$
    I'd start here: for each $n$, we can choose how many times $0$ occurs ,and how many times $1$ occurs. $-1$ then has to match $1$. So if $0$ is used $r$ times and $1$ is used $s$ times, $r=n-2s$.Then for each choice of $r$ and $s$ you've got a defined permutation problem, with $r, s$ and $s$ items..
    $endgroup$
    – timtfj
    Dec 29 '18 at 17:25
















  • 1




    $begingroup$
    Set up a recurrence perhaps.
    $endgroup$
    – paw88789
    Dec 29 '18 at 16:35












  • $begingroup$
    do you mean n each or a total of n?
    $endgroup$
    – player100
    Dec 29 '18 at 16:45










  • $begingroup$
    Sorry I meant that in total, the number of 0's, 1's and -1's is n
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 16:59










  • $begingroup$
    @MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
    $endgroup$
    – timtfj
    Dec 29 '18 at 17:11










  • $begingroup$
    I'd start here: for each $n$, we can choose how many times $0$ occurs ,and how many times $1$ occurs. $-1$ then has to match $1$. So if $0$ is used $r$ times and $1$ is used $s$ times, $r=n-2s$.Then for each choice of $r$ and $s$ you've got a defined permutation problem, with $r, s$ and $s$ items..
    $endgroup$
    – timtfj
    Dec 29 '18 at 17:25










1




1




$begingroup$
Set up a recurrence perhaps.
$endgroup$
– paw88789
Dec 29 '18 at 16:35






$begingroup$
Set up a recurrence perhaps.
$endgroup$
– paw88789
Dec 29 '18 at 16:35














$begingroup$
do you mean n each or a total of n?
$endgroup$
– player100
Dec 29 '18 at 16:45




$begingroup$
do you mean n each or a total of n?
$endgroup$
– player100
Dec 29 '18 at 16:45












$begingroup$
Sorry I meant that in total, the number of 0's, 1's and -1's is n
$endgroup$
– Lucio Tanzini
Dec 29 '18 at 16:59




$begingroup$
Sorry I meant that in total, the number of 0's, 1's and -1's is n
$endgroup$
– Lucio Tanzini
Dec 29 '18 at 16:59












$begingroup$
@MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
$endgroup$
– timtfj
Dec 29 '18 at 17:11




$begingroup$
@MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
$endgroup$
– timtfj
Dec 29 '18 at 17:11












$begingroup$
I'd start here: for each $n$, we can choose how many times $0$ occurs ,and how many times $1$ occurs. $-1$ then has to match $1$. So if $0$ is used $r$ times and $1$ is used $s$ times, $r=n-2s$.Then for each choice of $r$ and $s$ you've got a defined permutation problem, with $r, s$ and $s$ items..
$endgroup$
– timtfj
Dec 29 '18 at 17:25






$begingroup$
I'd start here: for each $n$, we can choose how many times $0$ occurs ,and how many times $1$ occurs. $-1$ then has to match $1$. So if $0$ is used $r$ times and $1$ is used $s$ times, $r=n-2s$.Then for each choice of $r$ and $s$ you've got a defined permutation problem, with $r, s$ and $s$ items..
$endgroup$
– timtfj
Dec 29 '18 at 17:25












2 Answers
2






active

oldest

votes


















4












$begingroup$

$s(n) = sum_{k=0}^{n/2} binom{n}{n-2k} cdot binom{2k}{k}$ The first term is the number of ways to arrange the zeroes, and then the second term arranges the parities. Now, we can simplify further:



$s(n) = sum_{k=0}^{n/2} frac{2k!}{2k!} cdot frac{n!}{(n-2k)!(k!)(k!)} = sum_{k=0}^{n/2} frac{n!}{(n-2k)!(k!)(k!)}$



Edit: put a few values into the OIES and came across trinomial coeffecients. In particular, s(n) is the n-th central trinomial coefficient, which has several closed forms, which you can find in the second link.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
    $endgroup$
    – timtfj
    Dec 29 '18 at 19:07










  • $begingroup$
    Whoops yeah I forgot a term
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:12










  • $begingroup$
    Yes, I got the same formula. Do you think It can be simplified into a closed formula?
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 19:31










  • $begingroup$
    @LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:37










  • $begingroup$
    Update: found closed form.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 20:43



















0












$begingroup$

The number of $0s, 1s$ and $-1s$ possible for a particular $n$ can be seen by the number of solutions to:



$$2p+q=n| p,q in Bbb Z^+$$



This can be done in $frac{n+1}{2}$ ways for odd $n$ and $frac{n+2}{2}$ ways for even $n$



You'll just need to account for positioning after this.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, but I'm afraid positioning is the main problem Indeed
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 17:34











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056013%2fin-how-many-ways-can-i-write-0-as-a-sum-of-n-0s-1s-textand-1s%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









4












$begingroup$

$s(n) = sum_{k=0}^{n/2} binom{n}{n-2k} cdot binom{2k}{k}$ The first term is the number of ways to arrange the zeroes, and then the second term arranges the parities. Now, we can simplify further:



$s(n) = sum_{k=0}^{n/2} frac{2k!}{2k!} cdot frac{n!}{(n-2k)!(k!)(k!)} = sum_{k=0}^{n/2} frac{n!}{(n-2k)!(k!)(k!)}$



Edit: put a few values into the OIES and came across trinomial coeffecients. In particular, s(n) is the n-th central trinomial coefficient, which has several closed forms, which you can find in the second link.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
    $endgroup$
    – timtfj
    Dec 29 '18 at 19:07










  • $begingroup$
    Whoops yeah I forgot a term
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:12










  • $begingroup$
    Yes, I got the same formula. Do you think It can be simplified into a closed formula?
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 19:31










  • $begingroup$
    @LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:37










  • $begingroup$
    Update: found closed form.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 20:43
















4












$begingroup$

$s(n) = sum_{k=0}^{n/2} binom{n}{n-2k} cdot binom{2k}{k}$ The first term is the number of ways to arrange the zeroes, and then the second term arranges the parities. Now, we can simplify further:



$s(n) = sum_{k=0}^{n/2} frac{2k!}{2k!} cdot frac{n!}{(n-2k)!(k!)(k!)} = sum_{k=0}^{n/2} frac{n!}{(n-2k)!(k!)(k!)}$



Edit: put a few values into the OIES and came across trinomial coeffecients. In particular, s(n) is the n-th central trinomial coefficient, which has several closed forms, which you can find in the second link.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
    $endgroup$
    – timtfj
    Dec 29 '18 at 19:07










  • $begingroup$
    Whoops yeah I forgot a term
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:12










  • $begingroup$
    Yes, I got the same formula. Do you think It can be simplified into a closed formula?
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 19:31










  • $begingroup$
    @LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:37










  • $begingroup$
    Update: found closed form.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 20:43














4












4








4





$begingroup$

$s(n) = sum_{k=0}^{n/2} binom{n}{n-2k} cdot binom{2k}{k}$ The first term is the number of ways to arrange the zeroes, and then the second term arranges the parities. Now, we can simplify further:



$s(n) = sum_{k=0}^{n/2} frac{2k!}{2k!} cdot frac{n!}{(n-2k)!(k!)(k!)} = sum_{k=0}^{n/2} frac{n!}{(n-2k)!(k!)(k!)}$



Edit: put a few values into the OIES and came across trinomial coeffecients. In particular, s(n) is the n-th central trinomial coefficient, which has several closed forms, which you can find in the second link.






share|cite|improve this answer











$endgroup$



$s(n) = sum_{k=0}^{n/2} binom{n}{n-2k} cdot binom{2k}{k}$ The first term is the number of ways to arrange the zeroes, and then the second term arranges the parities. Now, we can simplify further:



$s(n) = sum_{k=0}^{n/2} frac{2k!}{2k!} cdot frac{n!}{(n-2k)!(k!)(k!)} = sum_{k=0}^{n/2} frac{n!}{(n-2k)!(k!)(k!)}$



Edit: put a few values into the OIES and came across trinomial coeffecients. In particular, s(n) is the n-th central trinomial coefficient, which has several closed forms, which you can find in the second link.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 29 '18 at 21:48

























answered Dec 29 '18 at 18:53









Zachary HunterZachary Hunter

54111




54111












  • $begingroup$
    This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
    $endgroup$
    – timtfj
    Dec 29 '18 at 19:07










  • $begingroup$
    Whoops yeah I forgot a term
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:12










  • $begingroup$
    Yes, I got the same formula. Do you think It can be simplified into a closed formula?
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 19:31










  • $begingroup$
    @LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:37










  • $begingroup$
    Update: found closed form.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 20:43


















  • $begingroup$
    This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
    $endgroup$
    – timtfj
    Dec 29 '18 at 19:07










  • $begingroup$
    Whoops yeah I forgot a term
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:12










  • $begingroup$
    Yes, I got the same formula. Do you think It can be simplified into a closed formula?
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 19:31










  • $begingroup$
    @LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:37










  • $begingroup$
    Update: found closed form.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 20:43
















$begingroup$
This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
$endgroup$
– timtfj
Dec 29 '18 at 19:07




$begingroup$
This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
$endgroup$
– timtfj
Dec 29 '18 at 19:07












$begingroup$
Whoops yeah I forgot a term
$endgroup$
– Zachary Hunter
Dec 29 '18 at 19:12




$begingroup$
Whoops yeah I forgot a term
$endgroup$
– Zachary Hunter
Dec 29 '18 at 19:12












$begingroup$
Yes, I got the same formula. Do you think It can be simplified into a closed formula?
$endgroup$
– Lucio Tanzini
Dec 29 '18 at 19:31




$begingroup$
Yes, I got the same formula. Do you think It can be simplified into a closed formula?
$endgroup$
– Lucio Tanzini
Dec 29 '18 at 19:31












$begingroup$
@LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
$endgroup$
– Zachary Hunter
Dec 29 '18 at 19:37




$begingroup$
@LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
$endgroup$
– Zachary Hunter
Dec 29 '18 at 19:37












$begingroup$
Update: found closed form.
$endgroup$
– Zachary Hunter
Dec 29 '18 at 20:43




$begingroup$
Update: found closed form.
$endgroup$
– Zachary Hunter
Dec 29 '18 at 20:43











0












$begingroup$

The number of $0s, 1s$ and $-1s$ possible for a particular $n$ can be seen by the number of solutions to:



$$2p+q=n| p,q in Bbb Z^+$$



This can be done in $frac{n+1}{2}$ ways for odd $n$ and $frac{n+2}{2}$ ways for even $n$



You'll just need to account for positioning after this.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, but I'm afraid positioning is the main problem Indeed
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 17:34
















0












$begingroup$

The number of $0s, 1s$ and $-1s$ possible for a particular $n$ can be seen by the number of solutions to:



$$2p+q=n| p,q in Bbb Z^+$$



This can be done in $frac{n+1}{2}$ ways for odd $n$ and $frac{n+2}{2}$ ways for even $n$



You'll just need to account for positioning after this.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, but I'm afraid positioning is the main problem Indeed
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 17:34














0












0








0





$begingroup$

The number of $0s, 1s$ and $-1s$ possible for a particular $n$ can be seen by the number of solutions to:



$$2p+q=n| p,q in Bbb Z^+$$



This can be done in $frac{n+1}{2}$ ways for odd $n$ and $frac{n+2}{2}$ ways for even $n$



You'll just need to account for positioning after this.






share|cite|improve this answer











$endgroup$



The number of $0s, 1s$ and $-1s$ possible for a particular $n$ can be seen by the number of solutions to:



$$2p+q=n| p,q in Bbb Z^+$$



This can be done in $frac{n+1}{2}$ ways for odd $n$ and $frac{n+2}{2}$ ways for even $n$



You'll just need to account for positioning after this.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 29 '18 at 17:16

























answered Dec 29 '18 at 16:55









Rhys HughesRhys Hughes

5,1601427




5,1601427












  • $begingroup$
    Yes, but I'm afraid positioning is the main problem Indeed
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 17:34


















  • $begingroup$
    Yes, but I'm afraid positioning is the main problem Indeed
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 17:34
















$begingroup$
Yes, but I'm afraid positioning is the main problem Indeed
$endgroup$
– Lucio Tanzini
Dec 29 '18 at 17:34




$begingroup$
Yes, but I'm afraid positioning is the main problem Indeed
$endgroup$
– Lucio Tanzini
Dec 29 '18 at 17:34


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056013%2fin-how-many-ways-can-i-write-0-as-a-sum-of-n-0s-1s-textand-1s%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Human spaceflight

Image
Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
Read more

Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

Image
0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
Read more

張江高科駅

Image
張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
Read more