Mixing
How to solve this confusing permutation problem related to arrangement of books?
$begingroup$
Ms. Jones has 10 books that she is going to put on her bookshelf. Of these, 4 are
mathematics books, 3 are chemistry books, 2 are history books, and 1 is a language
book. Ms. Jones wants to arrange her books so that all the books dealing with the
same subject are together on the shelf. How many different arrangements are
possible?
I am completely lost. I know there are 4! 3! 2! 1! arrangements such that the mathematics books are
first in line, then the chemistry books, then the history books, and then the language
book.
But this is not the answer.
permutations combinations order-statistics
asked Oct 28 '15 at 8:42
RajSharmaRajSharma
373217
$endgroup$
add a comment |
$begingroup$
Ms. Jones has 10 books that she is going to put on her bookshelf. Of these, 4 are
mathematics books, 3 are chemistry books, 2 are history books, and 1 is a language
book. Ms. Jones wants to arrange her books so that all the books dealing with the
same subject are together on the shelf. How many different arrangements are
possible?
I am completely lost. I know there are 4! 3! 2! 1! arrangements such that the mathematics books are
first in line, then the chemistry books, then the history books, and then the language
book.
But this is not the answer.
permutations combinations order-statistics
asked Oct 28 '15 at 8:42
RajSharmaRajSharma
373217
$endgroup$
add a comment |
$begingroup$
Ms. Jones has 10 books that she is going to put on her bookshelf. Of these, 4 are
mathematics books, 3 are chemistry books, 2 are history books, and 1 is a language
book. Ms. Jones wants to arrange her books so that all the books dealing with the
same subject are together on the shelf. How many different arrangements are
possible?
I am completely lost. I know there are 4! 3! 2! 1! arrangements such that the mathematics books are
first in line, then the chemistry books, then the history books, and then the language
book.
But this is not the answer.
permutations combinations order-statistics
asked Oct 28 '15 at 8:42
RajSharmaRajSharma
373217
$endgroup$
Ms. Jones has 10 books that she is going to put on her bookshelf. Of these, 4 are
mathematics books, 3 are chemistry books, 2 are history books, and 1 is a language
book. Ms. Jones wants to arrange her books so that all the books dealing with the
same subject are together on the shelf. How many different arrangements are
possible?
I am completely lost. I know there are 4! 3! 2! 1! arrangements such that the mathematics books are
first in line, then the chemistry books, then the history books, and then the language
book.
But this is not the answer.
permutations combinations order-statistics
permutations combinations order-statistics
asked Oct 28 '15 at 8:42
RajSharmaRajSharma
373217
asked Oct 28 '15 at 8:42
RajSharmaRajSharma
373217
asked Oct 28 '15 at 8:42
RajSharmaRajSharma
373217
asked Oct 28 '15 at 8:42
RajSharmaRajSharma
373217
asked Oct 28 '15 at 8:42
RajSharmaRajSharma
373217
373217
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
you have already excellent answers so i will try to give you a nice way on how to see the problem.
Suppose that the books of the same category are all in a custody,so we have a custody for math,chemistry,history and language: $4$ custodies in total.
Now the question is: in how many ways can you arrange those 4 custodies?
The answer is clearly $4!$. Now what is left is the number of arrangements of books in the same custody,therefore we have $4!cdot 3! cdot 2! cdot 1!$ and you can do that for every arrangement of the $4$ custodies .,therefore you have finally $4!(4! cdot 3! cdot 2! cdot 1!) $
answered Oct 28 '15 at 9:08
NamelessNameless
745718
$endgroup$
add a comment |
$begingroup$
HINT:
$$underbrace{underbrace{M_1M_2M_3M_4}_{4!}midunderbrace{C_1C_2C_3}_{3!}midunderbrace{H_1H_2}_{2!}midunderbrace{L_1}_{1!}}_{4!}$$
answered Oct 28 '15 at 9:00
barak manosbarak manos
37.8k74199
$endgroup$
$begingroup$
This is more so a full-blown(but very helpful) answer than a hint
$endgroup$
– Cruncher
Oct 28 '15 at 17:45
$begingroup$
@Cruncher: Thanks :)
$endgroup$
– barak manos
Oct 28 '15 at 18:01
add a comment |
$begingroup$
You treat books of same subject as one object (since they have to be together) so you have $ 4!$ options. And then comes your part in which you permute books of same subject. So the final result is $4!4!3!2!1!$.
answered Oct 28 '15 at 8:49
user5216user5216
42627
$endgroup$
add a comment |
$begingroup$
You are almost there, but you need to multiply the arrangement for the categories. There are $4$ categories so the final answer should be $4!(4!3!2!1!)$.
answered Oct 28 '15 at 8:49
cr001cr001
7,382517
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
you have already excellent answers so i will try to give you a nice way on how to see the problem.
Suppose that the books of the same category are all in a custody,so we have a custody for math,chemistry,history and language: $4$ custodies in total.
Now the question is: in how many ways can you arrange those 4 custodies?
The answer is clearly $4!$. Now what is left is the number of arrangements of books in the same custody,therefore we have $4!cdot 3! cdot 2! cdot 1!$ and you can do that for every arrangement of the $4$ custodies .,therefore you have finally $4!(4! cdot 3! cdot 2! cdot 1!) $
answered Oct 28 '15 at 9:08
NamelessNameless
745718
$endgroup$
add a comment |
$begingroup$
you have already excellent answers so i will try to give you a nice way on how to see the problem.
Suppose that the books of the same category are all in a custody,so we have a custody for math,chemistry,history and language: $4$ custodies in total.
Now the question is: in how many ways can you arrange those 4 custodies?
The answer is clearly $4!$. Now what is left is the number of arrangements of books in the same custody,therefore we have $4!cdot 3! cdot 2! cdot 1!$ and you can do that for every arrangement of the $4$ custodies .,therefore you have finally $4!(4! cdot 3! cdot 2! cdot 1!) $
answered Oct 28 '15 at 9:08
NamelessNameless
745718
$endgroup$
add a comment |
$begingroup$
you have already excellent answers so i will try to give you a nice way on how to see the problem.
Suppose that the books of the same category are all in a custody,so we have a custody for math,chemistry,history and language: $4$ custodies in total.
Now the question is: in how many ways can you arrange those 4 custodies?
The answer is clearly $4!$. Now what is left is the number of arrangements of books in the same custody,therefore we have $4!cdot 3! cdot 2! cdot 1!$ and you can do that for every arrangement of the $4$ custodies .,therefore you have finally $4!(4! cdot 3! cdot 2! cdot 1!) $
answered Oct 28 '15 at 9:08
NamelessNameless
745718
$endgroup$
you have already excellent answers so i will try to give you a nice way on how to see the problem.
Suppose that the books of the same category are all in a custody,so we have a custody for math,chemistry,history and language: $4$ custodies in total.
Now the question is: in how many ways can you arrange those 4 custodies?
The answer is clearly $4!$. Now what is left is the number of arrangements of books in the same custody,therefore we have $4!cdot 3! cdot 2! cdot 1!$ and you can do that for every arrangement of the $4$ custodies .,therefore you have finally $4!(4! cdot 3! cdot 2! cdot 1!) $
answered Oct 28 '15 at 9:08
NamelessNameless
745718
edited Oct 28 '15 at 9:14
answered Oct 28 '15 at 9:08
NamelessNameless
745718
answered Oct 28 '15 at 9:08
NamelessNameless
745718
answered Oct 28 '15 at 9:08
NamelessNameless
745718
745718
add a comment |
add a comment |
$begingroup$
HINT:
$$underbrace{underbrace{M_1M_2M_3M_4}_{4!}midunderbrace{C_1C_2C_3}_{3!}midunderbrace{H_1H_2}_{2!}midunderbrace{L_1}_{1!}}_{4!}$$
answered Oct 28 '15 at 9:00
barak manosbarak manos
37.8k74199
$endgroup$
$begingroup$
This is more so a full-blown(but very helpful) answer than a hint
$endgroup$
– Cruncher
Oct 28 '15 at 17:45
$begingroup$
@Cruncher: Thanks :)
$endgroup$
– barak manos
Oct 28 '15 at 18:01
add a comment |
$begingroup$
HINT:
$$underbrace{underbrace{M_1M_2M_3M_4}_{4!}midunderbrace{C_1C_2C_3}_{3!}midunderbrace{H_1H_2}_{2!}midunderbrace{L_1}_{1!}}_{4!}$$
answered Oct 28 '15 at 9:00
barak manosbarak manos
37.8k74199
$endgroup$
$begingroup$
This is more so a full-blown(but very helpful) answer than a hint
$endgroup$
– Cruncher
Oct 28 '15 at 17:45
$begingroup$
@Cruncher: Thanks :)
$endgroup$
– barak manos
Oct 28 '15 at 18:01
add a comment |
$begingroup$
HINT:
$$underbrace{underbrace{M_1M_2M_3M_4}_{4!}midunderbrace{C_1C_2C_3}_{3!}midunderbrace{H_1H_2}_{2!}midunderbrace{L_1}_{1!}}_{4!}$$
answered Oct 28 '15 at 9:00
barak manosbarak manos
37.8k74199
$endgroup$
HINT:
$$underbrace{underbrace{M_1M_2M_3M_4}_{4!}midunderbrace{C_1C_2C_3}_{3!}midunderbrace{H_1H_2}_{2!}midunderbrace{L_1}_{1!}}_{4!}$$
answered Oct 28 '15 at 9:00
barak manosbarak manos
37.8k74199
edited Oct 28 '15 at 12:11
answered Oct 28 '15 at 9:00
barak manosbarak manos
37.8k74199
answered Oct 28 '15 at 9:00
barak manosbarak manos
37.8k74199
answered Oct 28 '15 at 9:00
barak manosbarak manos
37.8k74199
37.8k74199
$begingroup$
This is more so a full-blown(but very helpful) answer than a hint
$endgroup$
– Cruncher
Oct 28 '15 at 17:45
$begingroup$
@Cruncher: Thanks :)
$endgroup$
– barak manos
Oct 28 '15 at 18:01
add a comment |
$begingroup$
This is more so a full-blown(but very helpful) answer than a hint
$endgroup$
– Cruncher
Oct 28 '15 at 17:45
$begingroup$
@Cruncher: Thanks :)
$endgroup$
– barak manos
Oct 28 '15 at 18:01
$begingroup$
This is more so a full-blown(but very helpful) answer than a hint
$endgroup$
– Cruncher
Oct 28 '15 at 17:45
$begingroup$
This is more so a full-blown(but very helpful) answer than a hint
$endgroup$
– Cruncher
Oct 28 '15 at 17:45
$begingroup$
@Cruncher: Thanks :)
$endgroup$
– barak manos
Oct 28 '15 at 18:01
$begingroup$
@Cruncher: Thanks :)
$endgroup$
– barak manos
Oct 28 '15 at 18:01
add a comment |
$begingroup$
You treat books of same subject as one object (since they have to be together) so you have $ 4!$ options. And then comes your part in which you permute books of same subject. So the final result is $4!4!3!2!1!$.
answered Oct 28 '15 at 8:49
user5216user5216
42627
$endgroup$
add a comment |
$begingroup$
You treat books of same subject as one object (since they have to be together) so you have $ 4!$ options. And then comes your part in which you permute books of same subject. So the final result is $4!4!3!2!1!$.
answered Oct 28 '15 at 8:49
user5216user5216
42627
$endgroup$
add a comment |
$begingroup$
You treat books of same subject as one object (since they have to be together) so you have $ 4!$ options. And then comes your part in which you permute books of same subject. So the final result is $4!4!3!2!1!$.
answered Oct 28 '15 at 8:49
user5216user5216
42627
$endgroup$
You treat books of same subject as one object (since they have to be together) so you have $ 4!$ options. And then comes your part in which you permute books of same subject. So the final result is $4!4!3!2!1!$.
answered Oct 28 '15 at 8:49
user5216user5216
42627
answered Oct 28 '15 at 8:49
user5216user5216
42627
answered Oct 28 '15 at 8:49
user5216user5216
42627
answered Oct 28 '15 at 8:49
user5216user5216
42627
42627
add a comment |
add a comment |
$begingroup$
You are almost there, but you need to multiply the arrangement for the categories. There are $4$ categories so the final answer should be $4!(4!3!2!1!)$.
answered Oct 28 '15 at 8:49
cr001cr001
7,382517
$endgroup$
add a comment |
$begingroup$
You are almost there, but you need to multiply the arrangement for the categories. There are $4$ categories so the final answer should be $4!(4!3!2!1!)$.
answered Oct 28 '15 at 8:49
cr001cr001
7,382517
$endgroup$
add a comment |
$begingroup$
You are almost there, but you need to multiply the arrangement for the categories. There are $4$ categories so the final answer should be $4!(4!3!2!1!)$.
answered Oct 28 '15 at 8:49
cr001cr001
7,382517
$endgroup$
You are almost there, but you need to multiply the arrangement for the categories. There are $4$ categories so the final answer should be $4!(4!3!2!1!)$.
answered Oct 28 '15 at 8:49
cr001cr001
7,382517
answered Oct 28 '15 at 8:49
cr001cr001
7,382517
answered Oct 28 '15 at 8:49
cr001cr001
7,382517
answered Oct 28 '15 at 8:49
cr001cr001
7,382517
7,382517
add a comment |
add a comment |
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