Number of ways to break a stick

Multi tool use
$begingroup$
Number of ways of breaking a stick of length $ngeq1$ into $n$ pieces of unit length (at each step break one of the pieces with length $geq1$ into two pieces of integer lengths) is what?
I couldn’t figure out how to approach this question, can someone help me out?
combinatorics permutations combinations
$endgroup$
add a comment |
$begingroup$
Number of ways of breaking a stick of length $ngeq1$ into $n$ pieces of unit length (at each step break one of the pieces with length $geq1$ into two pieces of integer lengths) is what?
I couldn’t figure out how to approach this question, can someone help me out?
combinatorics permutations combinations
$endgroup$
1
$begingroup$
The points to break are clear, now we only need to take an order on them to get a breaking strategy...
$endgroup$
– dan_fulea
Jan 17 at 19:08
$begingroup$
Can you please elaborate
$endgroup$
– user601297
Jan 17 at 19:09
$begingroup$
Ok no need i get it now, answer should be (n-1)!, right ?
$endgroup$
– user601297
Jan 17 at 19:11
1
$begingroup$
Take a stick of length 5, so [0-1-2-3-4-5], now the breaking points are 1,2,3,4, take a permutation say 1234 $to$ 4132, then break first in 4, then in 1, ... So we count permutations finally...
$endgroup$
– dan_fulea
Jan 17 at 19:12
$begingroup$
Excellent, well done!
$endgroup$
– dan_fulea
Jan 17 at 19:12
add a comment |
$begingroup$
Number of ways of breaking a stick of length $ngeq1$ into $n$ pieces of unit length (at each step break one of the pieces with length $geq1$ into two pieces of integer lengths) is what?
I couldn’t figure out how to approach this question, can someone help me out?
combinatorics permutations combinations
$endgroup$
Number of ways of breaking a stick of length $ngeq1$ into $n$ pieces of unit length (at each step break one of the pieces with length $geq1$ into two pieces of integer lengths) is what?
I couldn’t figure out how to approach this question, can someone help me out?
combinatorics permutations combinations
combinatorics permutations combinations
asked Jan 17 at 19:05
user601297user601297
40019
40019
1
$begingroup$
The points to break are clear, now we only need to take an order on them to get a breaking strategy...
$endgroup$
– dan_fulea
Jan 17 at 19:08
$begingroup$
Can you please elaborate
$endgroup$
– user601297
Jan 17 at 19:09
$begingroup$
Ok no need i get it now, answer should be (n-1)!, right ?
$endgroup$
– user601297
Jan 17 at 19:11
1
$begingroup$
Take a stick of length 5, so [0-1-2-3-4-5], now the breaking points are 1,2,3,4, take a permutation say 1234 $to$ 4132, then break first in 4, then in 1, ... So we count permutations finally...
$endgroup$
– dan_fulea
Jan 17 at 19:12
$begingroup$
Excellent, well done!
$endgroup$
– dan_fulea
Jan 17 at 19:12
add a comment |
1
$begingroup$
The points to break are clear, now we only need to take an order on them to get a breaking strategy...
$endgroup$
– dan_fulea
Jan 17 at 19:08
$begingroup$
Can you please elaborate
$endgroup$
– user601297
Jan 17 at 19:09
$begingroup$
Ok no need i get it now, answer should be (n-1)!, right ?
$endgroup$
– user601297
Jan 17 at 19:11
1
$begingroup$
Take a stick of length 5, so [0-1-2-3-4-5], now the breaking points are 1,2,3,4, take a permutation say 1234 $to$ 4132, then break first in 4, then in 1, ... So we count permutations finally...
$endgroup$
– dan_fulea
Jan 17 at 19:12
$begingroup$
Excellent, well done!
$endgroup$
– dan_fulea
Jan 17 at 19:12
1
1
$begingroup$
The points to break are clear, now we only need to take an order on them to get a breaking strategy...
$endgroup$
– dan_fulea
Jan 17 at 19:08
$begingroup$
The points to break are clear, now we only need to take an order on them to get a breaking strategy...
$endgroup$
– dan_fulea
Jan 17 at 19:08
$begingroup$
Can you please elaborate
$endgroup$
– user601297
Jan 17 at 19:09
$begingroup$
Can you please elaborate
$endgroup$
– user601297
Jan 17 at 19:09
$begingroup$
Ok no need i get it now, answer should be (n-1)!, right ?
$endgroup$
– user601297
Jan 17 at 19:11
$begingroup$
Ok no need i get it now, answer should be (n-1)!, right ?
$endgroup$
– user601297
Jan 17 at 19:11
1
1
$begingroup$
Take a stick of length 5, so [0-1-2-3-4-5], now the breaking points are 1,2,3,4, take a permutation say 1234 $to$ 4132, then break first in 4, then in 1, ... So we count permutations finally...
$endgroup$
– dan_fulea
Jan 17 at 19:12
$begingroup$
Take a stick of length 5, so [0-1-2-3-4-5], now the breaking points are 1,2,3,4, take a permutation say 1234 $to$ 4132, then break first in 4, then in 1, ... So we count permutations finally...
$endgroup$
– dan_fulea
Jan 17 at 19:12
$begingroup$
Excellent, well done!
$endgroup$
– dan_fulea
Jan 17 at 19:12
$begingroup$
Excellent, well done!
$endgroup$
– dan_fulea
Jan 17 at 19:12
add a comment |
0
active
oldest
votes
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%2f3077385%2fnumber-of-ways-to-break-a-stick%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3077385%2fnumber-of-ways-to-break-a-stick%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
uX6hYk58,EU 1vhLScB cU 38qF0 cXjtElpso 4,zQUVEpVBDX JH,RaK8J3b5WFs0,7atKOpGrQrmG7OmIp
1
$begingroup$
The points to break are clear, now we only need to take an order on them to get a breaking strategy...
$endgroup$
– dan_fulea
Jan 17 at 19:08
$begingroup$
Can you please elaborate
$endgroup$
– user601297
Jan 17 at 19:09
$begingroup$
Ok no need i get it now, answer should be (n-1)!, right ?
$endgroup$
– user601297
Jan 17 at 19:11
1
$begingroup$
Take a stick of length 5, so [0-1-2-3-4-5], now the breaking points are 1,2,3,4, take a permutation say 1234 $to$ 4132, then break first in 4, then in 1, ... So we count permutations finally...
$endgroup$
– dan_fulea
Jan 17 at 19:12
$begingroup$
Excellent, well done!
$endgroup$
– dan_fulea
Jan 17 at 19:12