Permutations with constraint
Q. Find the number of different ways that $6$ boys and $4$ girls can stand in a line if
(i) all $6$ boys stand next to each other,
(ii) no girl stands next to another girl.
A. (i) There are $5$ "patterns", because I can start with $0$ to $4$ Gs, then put $6$ Bs, and then the remaining $4$ to $0$ Gs. For each pattern, $6!$ orders of B and $4!$ of G, so the answer is $5.6!.4! = 86400$.
(ii) As before, the answer is $k.6!.4!$, but it is not so easy to find $k$, the number of patterns. I can enumerate them as xGBxGBxGBxGx where each x represents $0$ to $3$ Bs, and the total of the xs is $3$. With some work, I find $k=35$. But there should be an easier way, I think.
combinatorics permutations
edited Nov 18 at 16:40
N. F. Taussig
43.5k93355
asked Nov 18 at 16:24
Rob625
111
add a comment |
Q. Find the number of different ways that $6$ boys and $4$ girls can stand in a line if
(i) all $6$ boys stand next to each other,
(ii) no girl stands next to another girl.
A. (i) There are $5$ "patterns", because I can start with $0$ to $4$ Gs, then put $6$ Bs, and then the remaining $4$ to $0$ Gs. For each pattern, $6!$ orders of B and $4!$ of G, so the answer is $5.6!.4! = 86400$.
(ii) As before, the answer is $k.6!.4!$, but it is not so easy to find $k$, the number of patterns. I can enumerate them as xGBxGBxGBxGx where each x represents $0$ to $3$ Bs, and the total of the xs is $3$. With some work, I find $k=35$. But there should be an easier way, I think.
combinatorics permutations
edited Nov 18 at 16:40
N. F. Taussig
43.5k93355
asked Nov 18 at 16:24
Rob625
111
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 18 at 16:41
add a comment |
Q. Find the number of different ways that $6$ boys and $4$ girls can stand in a line if
(i) all $6$ boys stand next to each other,
(ii) no girl stands next to another girl.
A. (i) There are $5$ "patterns", because I can start with $0$ to $4$ Gs, then put $6$ Bs, and then the remaining $4$ to $0$ Gs. For each pattern, $6!$ orders of B and $4!$ of G, so the answer is $5.6!.4! = 86400$.
(ii) As before, the answer is $k.6!.4!$, but it is not so easy to find $k$, the number of patterns. I can enumerate them as xGBxGBxGBxGx where each x represents $0$ to $3$ Bs, and the total of the xs is $3$. With some work, I find $k=35$. But there should be an easier way, I think.
combinatorics permutations
edited Nov 18 at 16:40
N. F. Taussig
43.5k93355
asked Nov 18 at 16:24
Rob625
111
Q. Find the number of different ways that $6$ boys and $4$ girls can stand in a line if
(i) all $6$ boys stand next to each other,
(ii) no girl stands next to another girl.
A. (i) There are $5$ "patterns", because I can start with $0$ to $4$ Gs, then put $6$ Bs, and then the remaining $4$ to $0$ Gs. For each pattern, $6!$ orders of B and $4!$ of G, so the answer is $5.6!.4! = 86400$.
(ii) As before, the answer is $k.6!.4!$, but it is not so easy to find $k$, the number of patterns. I can enumerate them as xGBxGBxGBxGx where each x represents $0$ to $3$ Bs, and the total of the xs is $3$. With some work, I find $k=35$. But there should be an easier way, I think.
combinatorics permutations
combinatorics permutations
edited Nov 18 at 16:40
N. F. Taussig
43.5k93355
asked Nov 18 at 16:24
Rob625
111
edited Nov 18 at 16:40
N. F. Taussig
43.5k93355
asked Nov 18 at 16:24
Rob625
111
edited Nov 18 at 16:40
N. F. Taussig
43.5k93355
edited Nov 18 at 16:40
N. F. Taussig
43.5k93355
edited Nov 18 at 16:40
N. F. Taussig
43.5k93355
43.5k93355
asked Nov 18 at 16:24
Rob625
111
asked Nov 18 at 16:24
Rob625
111
asked Nov 18 at 16:24
Rob625
111
111
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 18 at 16:41
add a comment |
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 18 at 16:41
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 18 at 16:41
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 18 at 16:41
add a comment |
2 Answers
2
active
oldest
votes
(ii)
You must find the number of sums $b_1+b_2+b_3+b_4+b_5=6$ where $b_1,b_5$ are nonnegative integers and $b_2,b_3,b_4$ are positive integers.
Here $b_1$ stands for the number of boys utmost left, $b_5$ for the number of boys utmost right and e.g. $b_2$ for the number of boys between the first and the second girl on the left.
Finding this number comes to the same as finding the number of sums $a_1+a_2+a_3+a_4+a_5=3$ where then $a_i$ are nonnegative numbers.
This arises if we substitute $b_1=a_1, b_5=a_5$ and $b_i=a_i+1$ for $i=2,3,4$.
Then with stars and bars we find $binom{3+4}4=35$ possibilities for that.
So the final number of possibilities equals $35times4!times6!$.
answered Nov 18 at 16:39
drhab
97.8k544129
add a comment |
The answers you have found are correct but there is simpler way to approach this.
(i) Consider the 6 boys to be a single person. Now the number of permutations will be $5!$.
However the boys themselves can be arranged in $6!$ ways. So, by principle of multiplication, the number of permutations if all the 6 boys stand together is $5!.6!=86400$
(ii) You need to visualize this in a different way to make things simpler.
Consider 6 boys standing with spaces between them i.e.
$ text{ X B X B X B X B X B X B X}$
where the $text{X}$s denote the empty spaces.
We need to place the girls in place of $text{X}$ which is the same as "no girl stands next to another girl".
This can be done in $P(7,4)$ or $binom{7}{4}.4!$ ways as there are 7 empty spaces or $text{X}$s and 4 girls.
Also, the boys themselves can be arranged in $6!$ ways.
Therefore, by principle of multiplication, the number of permutations that no girl stands next to another girl is $6!.binom{7}{4}.4!=604800$
answered Nov 18 at 16:55
Vaibhav
588
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%2f3003747%2fpermutations-with-constraint%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
(ii)
You must find the number of sums $b_1+b_2+b_3+b_4+b_5=6$ where $b_1,b_5$ are nonnegative integers and $b_2,b_3,b_4$ are positive integers.
Here $b_1$ stands for the number of boys utmost left, $b_5$ for the number of boys utmost right and e.g. $b_2$ for the number of boys between the first and the second girl on the left.
Finding this number comes to the same as finding the number of sums $a_1+a_2+a_3+a_4+a_5=3$ where then $a_i$ are nonnegative numbers.
This arises if we substitute $b_1=a_1, b_5=a_5$ and $b_i=a_i+1$ for $i=2,3,4$.
Then with stars and bars we find $binom{3+4}4=35$ possibilities for that.
So the final number of possibilities equals $35times4!times6!$.
answered Nov 18 at 16:39
drhab
97.8k544129
add a comment |
(ii)
You must find the number of sums $b_1+b_2+b_3+b_4+b_5=6$ where $b_1,b_5$ are nonnegative integers and $b_2,b_3,b_4$ are positive integers.
Here $b_1$ stands for the number of boys utmost left, $b_5$ for the number of boys utmost right and e.g. $b_2$ for the number of boys between the first and the second girl on the left.
Finding this number comes to the same as finding the number of sums $a_1+a_2+a_3+a_4+a_5=3$ where then $a_i$ are nonnegative numbers.
This arises if we substitute $b_1=a_1, b_5=a_5$ and $b_i=a_i+1$ for $i=2,3,4$.
Then with stars and bars we find $binom{3+4}4=35$ possibilities for that.
So the final number of possibilities equals $35times4!times6!$.
answered Nov 18 at 16:39
drhab
97.8k544129
add a comment |
(ii)
You must find the number of sums $b_1+b_2+b_3+b_4+b_5=6$ where $b_1,b_5$ are nonnegative integers and $b_2,b_3,b_4$ are positive integers.
Here $b_1$ stands for the number of boys utmost left, $b_5$ for the number of boys utmost right and e.g. $b_2$ for the number of boys between the first and the second girl on the left.
Finding this number comes to the same as finding the number of sums $a_1+a_2+a_3+a_4+a_5=3$ where then $a_i$ are nonnegative numbers.
This arises if we substitute $b_1=a_1, b_5=a_5$ and $b_i=a_i+1$ for $i=2,3,4$.
Then with stars and bars we find $binom{3+4}4=35$ possibilities for that.
So the final number of possibilities equals $35times4!times6!$.
answered Nov 18 at 16:39
drhab
97.8k544129
(ii)
You must find the number of sums $b_1+b_2+b_3+b_4+b_5=6$ where $b_1,b_5$ are nonnegative integers and $b_2,b_3,b_4$ are positive integers.
Here $b_1$ stands for the number of boys utmost left, $b_5$ for the number of boys utmost right and e.g. $b_2$ for the number of boys between the first and the second girl on the left.
Finding this number comes to the same as finding the number of sums $a_1+a_2+a_3+a_4+a_5=3$ where then $a_i$ are nonnegative numbers.
This arises if we substitute $b_1=a_1, b_5=a_5$ and $b_i=a_i+1$ for $i=2,3,4$.
Then with stars and bars we find $binom{3+4}4=35$ possibilities for that.
So the final number of possibilities equals $35times4!times6!$.
answered Nov 18 at 16:39
drhab
97.8k544129
answered Nov 18 at 16:39
drhab
97.8k544129
answered Nov 18 at 16:39
drhab
97.8k544129
answered Nov 18 at 16:39
drhab
97.8k544129
97.8k544129
add a comment |
add a comment |
The answers you have found are correct but there is simpler way to approach this.
(i) Consider the 6 boys to be a single person. Now the number of permutations will be $5!$.
However the boys themselves can be arranged in $6!$ ways. So, by principle of multiplication, the number of permutations if all the 6 boys stand together is $5!.6!=86400$
(ii) You need to visualize this in a different way to make things simpler.
Consider 6 boys standing with spaces between them i.e.
$ text{ X B X B X B X B X B X B X}$
where the $text{X}$s denote the empty spaces.
We need to place the girls in place of $text{X}$ which is the same as "no girl stands next to another girl".
This can be done in $P(7,4)$ or $binom{7}{4}.4!$ ways as there are 7 empty spaces or $text{X}$s and 4 girls.
Also, the boys themselves can be arranged in $6!$ ways.
Therefore, by principle of multiplication, the number of permutations that no girl stands next to another girl is $6!.binom{7}{4}.4!=604800$
answered Nov 18 at 16:55
Vaibhav
588
add a comment |
The answers you have found are correct but there is simpler way to approach this.
(i) Consider the 6 boys to be a single person. Now the number of permutations will be $5!$.
However the boys themselves can be arranged in $6!$ ways. So, by principle of multiplication, the number of permutations if all the 6 boys stand together is $5!.6!=86400$
(ii) You need to visualize this in a different way to make things simpler.
Consider 6 boys standing with spaces between them i.e.
$ text{ X B X B X B X B X B X B X}$
where the $text{X}$s denote the empty spaces.
We need to place the girls in place of $text{X}$ which is the same as "no girl stands next to another girl".
This can be done in $P(7,4)$ or $binom{7}{4}.4!$ ways as there are 7 empty spaces or $text{X}$s and 4 girls.
Also, the boys themselves can be arranged in $6!$ ways.
Therefore, by principle of multiplication, the number of permutations that no girl stands next to another girl is $6!.binom{7}{4}.4!=604800$
answered Nov 18 at 16:55
Vaibhav
588
add a comment |
The answers you have found are correct but there is simpler way to approach this.
(i) Consider the 6 boys to be a single person. Now the number of permutations will be $5!$.
However the boys themselves can be arranged in $6!$ ways. So, by principle of multiplication, the number of permutations if all the 6 boys stand together is $5!.6!=86400$
(ii) You need to visualize this in a different way to make things simpler.
Consider 6 boys standing with spaces between them i.e.
$ text{ X B X B X B X B X B X B X}$
where the $text{X}$s denote the empty spaces.
We need to place the girls in place of $text{X}$ which is the same as "no girl stands next to another girl".
This can be done in $P(7,4)$ or $binom{7}{4}.4!$ ways as there are 7 empty spaces or $text{X}$s and 4 girls.
Also, the boys themselves can be arranged in $6!$ ways.
Therefore, by principle of multiplication, the number of permutations that no girl stands next to another girl is $6!.binom{7}{4}.4!=604800$
answered Nov 18 at 16:55
Vaibhav
588
The answers you have found are correct but there is simpler way to approach this.
(i) Consider the 6 boys to be a single person. Now the number of permutations will be $5!$.
However the boys themselves can be arranged in $6!$ ways. So, by principle of multiplication, the number of permutations if all the 6 boys stand together is $5!.6!=86400$
(ii) You need to visualize this in a different way to make things simpler.
Consider 6 boys standing with spaces between them i.e.
$ text{ X B X B X B X B X B X B X}$
where the $text{X}$s denote the empty spaces.
We need to place the girls in place of $text{X}$ which is the same as "no girl stands next to another girl".
This can be done in $P(7,4)$ or $binom{7}{4}.4!$ ways as there are 7 empty spaces or $text{X}$s and 4 girls.
Also, the boys themselves can be arranged in $6!$ ways.
Therefore, by principle of multiplication, the number of permutations that no girl stands next to another girl is $6!.binom{7}{4}.4!=604800$
answered Nov 18 at 16:55
Vaibhav
588
answered Nov 18 at 16:55
Vaibhav
588
answered Nov 18 at 16:55
Vaibhav
588
answered Nov 18 at 16:55
Vaibhav
588
588
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3003747%2fpermutations-with-constraint%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
Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
– N. F. Taussig
Nov 18 at 16:41