Mixing
permutations indistinguishable objects and groups
$begingroup$
There is a group of 10 objects, 2 red, 3 blue and 5 green. If the 5 green objects should always be placed together, in how many ways we can put them on a line.
I did this:
As 5 places are occupied by the 5 green, that can be disposed in only 1 way as they are indistingushable, I did:
$ 5! / (3! * 2!) = (5 * 4 * 3 * 2 * 1) / *6 * 2) = 120/12=10 $
but I am not sure because the 5 green can be in any space.
I tried to draw it on paper and it came that 5 green on 10 spots can be arranged in 6 different ways.
So should I multiply 10 * 6 = 60?
I am not sure thought what is the formula for how to arrange the 5 on 10 spots
combinatorics permutations
asked Nov 12 '14 at 12:48
jsabjsab
43111
$endgroup$
add a comment |
$begingroup$
There is a group of 10 objects, 2 red, 3 blue and 5 green. If the 5 green objects should always be placed together, in how many ways we can put them on a line.
I did this:
As 5 places are occupied by the 5 green, that can be disposed in only 1 way as they are indistingushable, I did:
$ 5! / (3! * 2!) = (5 * 4 * 3 * 2 * 1) / *6 * 2) = 120/12=10 $
but I am not sure because the 5 green can be in any space.
I tried to draw it on paper and it came that 5 green on 10 spots can be arranged in 6 different ways.
So should I multiply 10 * 6 = 60?
I am not sure thought what is the formula for how to arrange the 5 on 10 spots
combinatorics permutations
asked Nov 12 '14 at 12:48
jsabjsab
43111
$endgroup$
add a comment |
$begingroup$
There is a group of 10 objects, 2 red, 3 blue and 5 green. If the 5 green objects should always be placed together, in how many ways we can put them on a line.
I did this:
As 5 places are occupied by the 5 green, that can be disposed in only 1 way as they are indistingushable, I did:
$ 5! / (3! * 2!) = (5 * 4 * 3 * 2 * 1) / *6 * 2) = 120/12=10 $
but I am not sure because the 5 green can be in any space.
I tried to draw it on paper and it came that 5 green on 10 spots can be arranged in 6 different ways.
So should I multiply 10 * 6 = 60?
I am not sure thought what is the formula for how to arrange the 5 on 10 spots
combinatorics permutations
asked Nov 12 '14 at 12:48
jsabjsab
43111
$endgroup$
There is a group of 10 objects, 2 red, 3 blue and 5 green. If the 5 green objects should always be placed together, in how many ways we can put them on a line.
I did this:
As 5 places are occupied by the 5 green, that can be disposed in only 1 way as they are indistingushable, I did:
$ 5! / (3! * 2!) = (5 * 4 * 3 * 2 * 1) / *6 * 2) = 120/12=10 $
but I am not sure because the 5 green can be in any space.
I tried to draw it on paper and it came that 5 green on 10 spots can be arranged in 6 different ways.
So should I multiply 10 * 6 = 60?
I am not sure thought what is the formula for how to arrange the 5 on 10 spots
combinatorics permutations
combinatorics permutations
asked Nov 12 '14 at 12:48
jsabjsab
43111
asked Nov 12 '14 at 12:48
jsabjsab
43111
asked Nov 12 '14 at 12:48
jsabjsab
43111
asked Nov 12 '14 at 12:48
jsabjsab
43111
asked Nov 12 '14 at 12:48
jsabjsab
43111
43111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your answer seems to be correct. $5$ green objects may be placed in $6$ ways, as you correctly found. After that we have $5$ places for $5$ objects of two types. This problem can be reduced to a simple permutation problem: we can just find in how many ways blue or red objects can be places and then fill the rest places with the objects of another color.
Thus, we have $binom 53$ (if we consider permutations of blue) or $binom 52$ (if we take red) for every fixed position of green objects. Anyway, $binom nk = binom {n}{n-k}$, so we can say that total number of permutations is $6 * binom 52 = 6 * frac{5!}{2!3!} = 60.$
answered Nov 12 '14 at 13:03
Andrei RykhalskiAndrei Rykhalski
1,237614
$endgroup$
$begingroup$
Thanks Andrei, but how I get to 6 with a formula? because I draw it on paper, but not sure how to get it..
$endgroup$
– jsab
Nov 12 '14 at 13:08
add a comment |
$begingroup$
The easiest way I could see this is:
This is the same task as arranging 6 objects: 2 red, 3 blue and 1 block-of-five-greens.
If you have $a_i$ objects of type $i$, there are $frac{(sum{a_i})!}{prod{(a_i!)}}$ permutations.
Hence you are correct. There are $frac{6!}{3! 2! 1!} = 60$ arrangements.
answered Jan 30 at 16:41
drjpizzledrjpizzle
1712
$endgroup$
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%2f1018451%2fpermutations-indistinguishable-objects-and-groups%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
$begingroup$
Your answer seems to be correct. $5$ green objects may be placed in $6$ ways, as you correctly found. After that we have $5$ places for $5$ objects of two types. This problem can be reduced to a simple permutation problem: we can just find in how many ways blue or red objects can be places and then fill the rest places with the objects of another color.
Thus, we have $binom 53$ (if we consider permutations of blue) or $binom 52$ (if we take red) for every fixed position of green objects. Anyway, $binom nk = binom {n}{n-k}$, so we can say that total number of permutations is $6 * binom 52 = 6 * frac{5!}{2!3!} = 60.$
answered Nov 12 '14 at 13:03
Andrei RykhalskiAndrei Rykhalski
1,237614
$endgroup$
$begingroup$
Thanks Andrei, but how I get to 6 with a formula? because I draw it on paper, but not sure how to get it..
$endgroup$
– jsab
Nov 12 '14 at 13:08
add a comment |
$begingroup$
Your answer seems to be correct. $5$ green objects may be placed in $6$ ways, as you correctly found. After that we have $5$ places for $5$ objects of two types. This problem can be reduced to a simple permutation problem: we can just find in how many ways blue or red objects can be places and then fill the rest places with the objects of another color.
Thus, we have $binom 53$ (if we consider permutations of blue) or $binom 52$ (if we take red) for every fixed position of green objects. Anyway, $binom nk = binom {n}{n-k}$, so we can say that total number of permutations is $6 * binom 52 = 6 * frac{5!}{2!3!} = 60.$
answered Nov 12 '14 at 13:03
Andrei RykhalskiAndrei Rykhalski
1,237614
$endgroup$
$begingroup$
Thanks Andrei, but how I get to 6 with a formula? because I draw it on paper, but not sure how to get it..
$endgroup$
– jsab
Nov 12 '14 at 13:08
add a comment |
$begingroup$
Your answer seems to be correct. $5$ green objects may be placed in $6$ ways, as you correctly found. After that we have $5$ places for $5$ objects of two types. This problem can be reduced to a simple permutation problem: we can just find in how many ways blue or red objects can be places and then fill the rest places with the objects of another color.
Thus, we have $binom 53$ (if we consider permutations of blue) or $binom 52$ (if we take red) for every fixed position of green objects. Anyway, $binom nk = binom {n}{n-k}$, so we can say that total number of permutations is $6 * binom 52 = 6 * frac{5!}{2!3!} = 60.$
answered Nov 12 '14 at 13:03
Andrei RykhalskiAndrei Rykhalski
1,237614
$endgroup$
Your answer seems to be correct. $5$ green objects may be placed in $6$ ways, as you correctly found. After that we have $5$ places for $5$ objects of two types. This problem can be reduced to a simple permutation problem: we can just find in how many ways blue or red objects can be places and then fill the rest places with the objects of another color.
Thus, we have $binom 53$ (if we consider permutations of blue) or $binom 52$ (if we take red) for every fixed position of green objects. Anyway, $binom nk = binom {n}{n-k}$, so we can say that total number of permutations is $6 * binom 52 = 6 * frac{5!}{2!3!} = 60.$
answered Nov 12 '14 at 13:03
Andrei RykhalskiAndrei Rykhalski
1,237614
answered Nov 12 '14 at 13:03
Andrei RykhalskiAndrei Rykhalski
1,237614
answered Nov 12 '14 at 13:03
Andrei RykhalskiAndrei Rykhalski
1,237614
answered Nov 12 '14 at 13:03
Andrei RykhalskiAndrei Rykhalski
1,237614
1,237614
$begingroup$
Thanks Andrei, but how I get to 6 with a formula? because I draw it on paper, but not sure how to get it..
$endgroup$
– jsab
Nov 12 '14 at 13:08
add a comment |
$begingroup$
Thanks Andrei, but how I get to 6 with a formula? because I draw it on paper, but not sure how to get it..
$endgroup$
– jsab
Nov 12 '14 at 13:08
$begingroup$
Thanks Andrei, but how I get to 6 with a formula? because I draw it on paper, but not sure how to get it..
$endgroup$
– jsab
Nov 12 '14 at 13:08
$begingroup$
Thanks Andrei, but how I get to 6 with a formula? because I draw it on paper, but not sure how to get it..
$endgroup$
– jsab
Nov 12 '14 at 13:08
add a comment |
$begingroup$
The easiest way I could see this is:
This is the same task as arranging 6 objects: 2 red, 3 blue and 1 block-of-five-greens.
If you have $a_i$ objects of type $i$, there are $frac{(sum{a_i})!}{prod{(a_i!)}}$ permutations.
Hence you are correct. There are $frac{6!}{3! 2! 1!} = 60$ arrangements.
answered Jan 30 at 16:41
drjpizzledrjpizzle
1712
$endgroup$
add a comment |
$begingroup$
The easiest way I could see this is:
This is the same task as arranging 6 objects: 2 red, 3 blue and 1 block-of-five-greens.
If you have $a_i$ objects of type $i$, there are $frac{(sum{a_i})!}{prod{(a_i!)}}$ permutations.
Hence you are correct. There are $frac{6!}{3! 2! 1!} = 60$ arrangements.
answered Jan 30 at 16:41
drjpizzledrjpizzle
1712
$endgroup$
add a comment |
$begingroup$
The easiest way I could see this is:
This is the same task as arranging 6 objects: 2 red, 3 blue and 1 block-of-five-greens.
If you have $a_i$ objects of type $i$, there are $frac{(sum{a_i})!}{prod{(a_i!)}}$ permutations.
Hence you are correct. There are $frac{6!}{3! 2! 1!} = 60$ arrangements.
answered Jan 30 at 16:41
drjpizzledrjpizzle
1712
$endgroup$
The easiest way I could see this is:
This is the same task as arranging 6 objects: 2 red, 3 blue and 1 block-of-five-greens.
If you have $a_i$ objects of type $i$, there are $frac{(sum{a_i})!}{prod{(a_i!)}}$ permutations.
Hence you are correct. There are $frac{6!}{3! 2! 1!} = 60$ arrangements.
answered Jan 30 at 16:41
drjpizzledrjpizzle
1712
answered Jan 30 at 16:41
drjpizzledrjpizzle
1712
answered Jan 30 at 16:41
drjpizzledrjpizzle
1712
answered Jan 30 at 16:41
drjpizzledrjpizzle
1712
1712
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%2f1018451%2fpermutations-indistinguishable-objects-and-groups%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
