Dtjytyk

"Oh no!" in Latin

How to make money from a browser who sees 5 seconds into the future of any web page?

Storage of electrolytic capacitors - how long?

Is there a reason to prefer HFS+ over APFS for disk images in High Sierra and/or Mojave?

In One Punch Man, is King actually weak?

Given this phrasing in the lease, when should I pay my rent?

Typing CO_2 easily

Should I warn a new PhD Student?

Sound waves in different octaves

Would this string work as string?

Origin of pigs as a species

Deciphering cause of death?

Are Captain Marvel's powers affected by Thanos breaking the Tesseract and claiming the stone?

Did I make a mistake by ccing email to boss to others?

Language involving irrational number is not a CFL

Limit max CPU usage SQL SERVER with WSRM

Ways of geometrical multiplication

Echo with obfuscation

What should be the ideal length of sentences in a blog post for ease of reading?

Can I say "fingers" when referring to toes?

How to I force windows to use a specific version of SQLCMD?

Why the various definitions of the thin space ,?

Why is the Sun approximated as a black body at ~ 5800 K?

Mimic lecturing on blackboard, facing audience

How many arrangements of six books where two blue books must be next to each other?

Arranging identical blue and red bottles such that there are no 2 red bottles next to each otherHow many ways can we arrange 7 books, including 2 math books and 1 physics book, with the math books next to each other and left of the physics book?In how many ways can 9 cars be parked so that there are never two red cars next to each other?How many ways can six 3’s and four 2’s be arranged in a row so that 2’s are always apart?In how many arrangements of the letters of the word INTERMEDIATE 2 vowels never come together?Number of arrangements of four blue, three green, and two red balls in which no two blue balls are adjacentPermutations- Restriction on two items placed next to each other.How many ways can six distinct black books and four distinct red books be arranged on a shelf if no two red books are adjacent?How many ways can you arrange books in a shelf?Seating six boys and four girls around a table so that no two girls can sit next to each other

1

$begingroup$

There four black books and two blue books. How many ways are there of
arranging the six books in a line such that the two blue books must
be directly next to one another?

My book gives the following answer:

$$5 times 2 times 4 times 3 times 2 times 1 $$

If I understand this correctly, we lump the two blue books together so that we are only considering the arrangement of $5$ different books, but we must multiply by $2$ since there are $2$ blue books, so there are $2!$ ways to arrange them. Does that sound correct? If there were $3$ blue books, would the answer be:

$$5 times 3 times 4 times 3 times 2 times 1 ?$$

combinatorics

share|cite|improve this question

edited Mar 12 at 23:05

Joseph Martin

690317

asked Mar 12 at 21:35

ZakuZaku

1678

$endgroup$

add a comment | 

1

$begingroup$

There four black books and two blue books. How many ways are there of
arranging the six books in a line such that the two blue books must
be directly next to one another?

My book gives the following answer:

$$5 times 2 times 4 times 3 times 2 times 1 $$

If I understand this correctly, we lump the two blue books together so that we are only considering the arrangement of $5$ different books, but we must multiply by $2$ since there are $2$ blue books, so there are $2!$ ways to arrange them. Does that sound correct? If there were $3$ blue books, would the answer be:

$$5 times 3 times 4 times 3 times 2 times 1 ?$$

combinatorics

share|cite|improve this question

edited Mar 12 at 23:05

Joseph Martin

690317

asked Mar 12 at 21:35

ZakuZaku

1678

$endgroup$

add a comment | 

$begingroup$

There four black books and two blue books. How many ways are there of
arranging the six books in a line such that the two blue books must
be directly next to one another?

My book gives the following answer:

$$5 times 2 times 4 times 3 times 2 times 1 $$

If I understand this correctly, we lump the two blue books together so that we are only considering the arrangement of $5$ different books, but we must multiply by $2$ since there are $2$ blue books, so there are $2!$ ways to arrange them. Does that sound correct? If there were $3$ blue books, would the answer be:

$$5 times 3 times 4 times 3 times 2 times 1 ?$$

combinatorics

share|cite|improve this question

edited Mar 12 at 23:05

Joseph Martin

690317

asked Mar 12 at 21:35

ZakuZaku

1678

$endgroup$

share|cite|improve this question

edited Mar 12 at 23:05

Joseph Martin

690317

asked Mar 12 at 21:35

ZakuZaku

1678

share|cite|improve this question

edited Mar 12 at 23:05

Joseph Martin

690317

asked Mar 12 at 21:35

ZakuZaku

1678

edited Mar 12 at 23:05

Joseph Martin

690317

edited Mar 12 at 23:05

Joseph Martin

690317

asked Mar 12 at 21:35

ZakuZaku

1678

asked Mar 12 at 21:35

ZakuZaku

1678

1 Answer
1

active

oldest

votes

1

$begingroup$

Not quite. There are 3! = 6 ways of arranging the blue books in your second example, so the answer would be $$5 times 4 times 3 times 2 times 1 times 3 times 2 times 1 = 720.$$

We can state it generally as:

There $k$ black books and $m$ blue books. How many ways are there of
arranging the $k+m$ books in a line such that the $m$ blue books must
all be directly next to one another?

Glue all of $m$ blue books together so there are effectively $k+1$ things to arrange, and we know that there are $(k+1)!$ ways to do that. We must, however, have this for each of the $m!$ ways there are to arrange the $m$ blue books. Therefore, our answer is $(k+1)! m!$.

share|cite|improve this answer

answered Mar 12 at 21:57

Joseph MartinJoseph Martin

690317

$endgroup$

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
});


}
});

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3145720%2fhow-many-arrangements-of-six-books-where-two-blue-books-must-be-next-to-each-oth%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

Not quite. There are 3! = 6 ways of arranging the blue books in your second example, so the answer would be $$5 times 4 times 3 times 2 times 1 times 3 times 2 times 1 = 720.$$

We can state it generally as:

There $k$ black books and $m$ blue books. How many ways are there of
arranging the $k+m$ books in a line such that the $m$ blue books must
all be directly next to one another?

Glue all of $m$ blue books together so there are effectively $k+1$ things to arrange, and we know that there are $(k+1)!$ ways to do that. We must, however, have this for each of the $m!$ ways there are to arrange the $m$ blue books. Therefore, our answer is $(k+1)! m!$.

share|cite|improve this answer

answered Mar 12 at 21:57

Joseph MartinJoseph Martin

690317

$endgroup$

add a comment | 

1

$begingroup$

Not quite. There are 3! = 6 ways of arranging the blue books in your second example, so the answer would be $$5 times 4 times 3 times 2 times 1 times 3 times 2 times 1 = 720.$$

We can state it generally as:

There $k$ black books and $m$ blue books. How many ways are there of
arranging the $k+m$ books in a line such that the $m$ blue books must
all be directly next to one another?

Glue all of $m$ blue books together so there are effectively $k+1$ things to arrange, and we know that there are $(k+1)!$ ways to do that. We must, however, have this for each of the $m!$ ways there are to arrange the $m$ blue books. Therefore, our answer is $(k+1)! m!$.

share|cite|improve this answer

answered Mar 12 at 21:57

Joseph MartinJoseph Martin

690317

$endgroup$

add a comment | 

$begingroup$

Not quite. There are 3! = 6 ways of arranging the blue books in your second example, so the answer would be $$5 times 4 times 3 times 2 times 1 times 3 times 2 times 1 = 720.$$

We can state it generally as:

There $k$ black books and $m$ blue books. How many ways are there of
arranging the $k+m$ books in a line such that the $m$ blue books must
all be directly next to one another?

Glue all of $m$ blue books together so there are effectively $k+1$ things to arrange, and we know that there are $(k+1)!$ ways to do that. We must, however, have this for each of the $m!$ ways there are to arrange the $m$ blue books. Therefore, our answer is $(k+1)! m!$.

share|cite|improve this answer

answered Mar 12 at 21:57

Joseph MartinJoseph Martin

690317

$endgroup$

share|cite|improve this answer

answered Mar 12 at 21:57

Joseph MartinJoseph Martin

690317

answered Mar 12 at 21:57

Joseph MartinJoseph Martin

690317

answered Mar 12 at 21:57

Joseph MartinJoseph Martin

690317