Mixing
How many ways can you arrange books in a shelf?
$begingroup$
There are 6 English books, 4 Science books, 7 magazines, and 3 Mathematics books. In how many ways can you arrange the shelf if:
a) English and Science books are indistinct?
b) English books should be together?
Pls someone help me on this one. Thank you!
EDIT: I actually have an initial answer. For a, $(10!)/(6! 4!)$ ways for the English and Science books, then multiply to $10!$ (ways for the others)? Is this correct? I'm actually not sure if I understand the restriction in a correctly.
combinatorics permutations combinations
asked Jan 19 at 7:23
cluelessclueless
266
$endgroup$
add a comment |
$begingroup$
There are 6 English books, 4 Science books, 7 magazines, and 3 Mathematics books. In how many ways can you arrange the shelf if:
a) English and Science books are indistinct?
b) English books should be together?
Pls someone help me on this one. Thank you!
EDIT: I actually have an initial answer. For a, $(10!)/(6! 4!)$ ways for the English and Science books, then multiply to $10!$ (ways for the others)? Is this correct? I'm actually not sure if I understand the restriction in a correctly.
combinatorics permutations combinations
asked Jan 19 at 7:23
cluelessclueless
266
$endgroup$
1
$begingroup$
Have you seen similar questions on the site? By similar, I refer to counting the arrangement of objects of different types. Look at them, and reference any that you find similar in your post above.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:24
1
$begingroup$
Please read the related questions at right, then make an attempt at solving the problem yourself.
$endgroup$
– N. F. Taussig
Jan 19 at 10:45
add a comment |
$begingroup$
There are 6 English books, 4 Science books, 7 magazines, and 3 Mathematics books. In how many ways can you arrange the shelf if:
a) English and Science books are indistinct?
b) English books should be together?
Pls someone help me on this one. Thank you!
EDIT: I actually have an initial answer. For a, $(10!)/(6! 4!)$ ways for the English and Science books, then multiply to $10!$ (ways for the others)? Is this correct? I'm actually not sure if I understand the restriction in a correctly.
combinatorics permutations combinations
asked Jan 19 at 7:23
cluelessclueless
266
$endgroup$
There are 6 English books, 4 Science books, 7 magazines, and 3 Mathematics books. In how many ways can you arrange the shelf if:
a) English and Science books are indistinct?
b) English books should be together?
Pls someone help me on this one. Thank you!
EDIT: I actually have an initial answer. For a, $(10!)/(6! 4!)$ ways for the English and Science books, then multiply to $10!$ (ways for the others)? Is this correct? I'm actually not sure if I understand the restriction in a correctly.
combinatorics permutations combinations
combinatorics permutations combinations
asked Jan 19 at 7:23
cluelessclueless
266
asked Jan 19 at 7:23
cluelessclueless
266
edited Jan 19 at 11:51
clueless
asked Jan 19 at 7:23
cluelessclueless
266
asked Jan 19 at 7:23
cluelessclueless
266
asked Jan 19 at 7:23
cluelessclueless
266
266
1
$begingroup$
Have you seen similar questions on the site? By similar, I refer to counting the arrangement of objects of different types. Look at them, and reference any that you find similar in your post above.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:24
1
$begingroup$
Please read the related questions at right, then make an attempt at solving the problem yourself.
$endgroup$
– N. F. Taussig
Jan 19 at 10:45
add a comment |
1
$begingroup$
Have you seen similar questions on the site? By similar, I refer to counting the arrangement of objects of different types. Look at them, and reference any that you find similar in your post above.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:24
1
$begingroup$
Please read the related questions at right, then make an attempt at solving the problem yourself.
$endgroup$
– N. F. Taussig
Jan 19 at 10:45
1
1
$begingroup$
Have you seen similar questions on the site? By similar, I refer to counting the arrangement of objects of different types. Look at them, and reference any that you find similar in your post above.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:24
$begingroup$
Have you seen similar questions on the site? By similar, I refer to counting the arrangement of objects of different types. Look at them, and reference any that you find similar in your post above.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:24
1
1
$begingroup$
Please read the related questions at right, then make an attempt at solving the problem yourself.
$endgroup$
– N. F. Taussig
Jan 19 at 10:45
$begingroup$
Please read the related questions at right, then make an attempt at solving the problem yourself.
$endgroup$
– N. F. Taussig
Jan 19 at 10:45
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In how many ways can $6$ English books, $4$ science books, $7$ magazines, and $3$ mathematics books be arranged on a shelf if English books are indistinguishable and science books are indistinguishable?
We have a total of $6 + 4 + 7 + 3 = 20$ books. Choose six of the $20$ positions for the English books and four of the remaining $14$ positions for the science books. The remaining ten positions can be filled with books and magazines in $10!$ ways.
$$binom{20}{6}binom{14}{4}10! = frac{20!}{6!14!} cdot binom{14!}{4!10!} cdot 10! = frac{20!}{6!4!}$$
In your attempt, you did not take into account the total number of positions on the shelf.
In how many ways can $6$ English books, $4$ science books, $7$ magazines, and $3$ mathematics books be arranged on a shelf if English books should be together?
If all the books are intended to be distinct (switching the order of the questions would have made this clearer), treat the English books as a single object, so we have $1 + 4 + 7 + 3 = 15$ objects to arrange. Then multiply by the number of ways of arranging the six English books within the block of English books.
If we are still supposed to treat the English books as being indistinguishable and the science books as being indistinguishable, choose six of the $15$ positions for the science books, one of the remaining $8$ positions for the block of English books, then arrange the magazines and mathematics books in the remaining positions.
I believe the first of these two interpretations is intended, but I would have reversed the order of the questions to make that clear.
answered Jan 19 at 13:39
N. F. TaussigN. F. Taussig
44.5k103357
$endgroup$
$begingroup$
Oh right. But for a, is it also the same as using the formula $n!/(n_1 ! times n_2 ! times n_3 ! times ... times n_{12}!)$ where $n_1=6, n_2=4$, and the others are $1!$ (since they're all distinct)?
$endgroup$
– clueless
Jan 19 at 15:52
$begingroup$
And for b, the answer should be $(15!)(6!)$. Is this correct? (assuming the first interpretation is intended)
$endgroup$
– clueless
Jan 19 at 16:01
$begingroup$
@clueless You are correct on both counts.
$endgroup$
– N. F. Taussig
Jan 19 at 16:04
1
$begingroup$
Thank you so much!
$endgroup$
– clueless
Jan 19 at 16:07
add a comment |
$begingroup$
The number of different permutations of $n$ objects, where $n_{1}$ are of one kind, $n_{2}$ are a different kind $dots$ and there are $k$ different kinds is:
$$
frac{n!}{n_{1}! times n_{2}! times dots times n_{k}!}
$$
answered Jan 19 at 13:09
C_RichmondC_Richmond
863
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
oldest
votes
$begingroup$
In how many ways can $6$ English books, $4$ science books, $7$ magazines, and $3$ mathematics books be arranged on a shelf if English books are indistinguishable and science books are indistinguishable?
We have a total of $6 + 4 + 7 + 3 = 20$ books. Choose six of the $20$ positions for the English books and four of the remaining $14$ positions for the science books. The remaining ten positions can be filled with books and magazines in $10!$ ways.
$$binom{20}{6}binom{14}{4}10! = frac{20!}{6!14!} cdot binom{14!}{4!10!} cdot 10! = frac{20!}{6!4!}$$
In your attempt, you did not take into account the total number of positions on the shelf.
In how many ways can $6$ English books, $4$ science books, $7$ magazines, and $3$ mathematics books be arranged on a shelf if English books should be together?
If all the books are intended to be distinct (switching the order of the questions would have made this clearer), treat the English books as a single object, so we have $1 + 4 + 7 + 3 = 15$ objects to arrange. Then multiply by the number of ways of arranging the six English books within the block of English books.
If we are still supposed to treat the English books as being indistinguishable and the science books as being indistinguishable, choose six of the $15$ positions for the science books, one of the remaining $8$ positions for the block of English books, then arrange the magazines and mathematics books in the remaining positions.
I believe the first of these two interpretations is intended, but I would have reversed the order of the questions to make that clear.
answered Jan 19 at 13:39
N. F. TaussigN. F. Taussig
44.5k103357
$endgroup$
$begingroup$
Oh right. But for a, is it also the same as using the formula $n!/(n_1 ! times n_2 ! times n_3 ! times ... times n_{12}!)$ where $n_1=6, n_2=4$, and the others are $1!$ (since they're all distinct)?
$endgroup$
– clueless
Jan 19 at 15:52
$begingroup$
And for b, the answer should be $(15!)(6!)$. Is this correct? (assuming the first interpretation is intended)
$endgroup$
– clueless
Jan 19 at 16:01
$begingroup$
@clueless You are correct on both counts.
$endgroup$
– N. F. Taussig
Jan 19 at 16:04
1
$begingroup$
Thank you so much!
$endgroup$
– clueless
Jan 19 at 16:07
add a comment |
$begingroup$
In how many ways can $6$ English books, $4$ science books, $7$ magazines, and $3$ mathematics books be arranged on a shelf if English books are indistinguishable and science books are indistinguishable?
We have a total of $6 + 4 + 7 + 3 = 20$ books. Choose six of the $20$ positions for the English books and four of the remaining $14$ positions for the science books. The remaining ten positions can be filled with books and magazines in $10!$ ways.
$$binom{20}{6}binom{14}{4}10! = frac{20!}{6!14!} cdot binom{14!}{4!10!} cdot 10! = frac{20!}{6!4!}$$
In your attempt, you did not take into account the total number of positions on the shelf.
In how many ways can $6$ English books, $4$ science books, $7$ magazines, and $3$ mathematics books be arranged on a shelf if English books should be together?
If all the books are intended to be distinct (switching the order of the questions would have made this clearer), treat the English books as a single object, so we have $1 + 4 + 7 + 3 = 15$ objects to arrange. Then multiply by the number of ways of arranging the six English books within the block of English books.
If we are still supposed to treat the English books as being indistinguishable and the science books as being indistinguishable, choose six of the $15$ positions for the science books, one of the remaining $8$ positions for the block of English books, then arrange the magazines and mathematics books in the remaining positions.
I believe the first of these two interpretations is intended, but I would have reversed the order of the questions to make that clear.
answered Jan 19 at 13:39
N. F. TaussigN. F. Taussig
44.5k103357
$endgroup$
$begingroup$
Oh right. But for a, is it also the same as using the formula $n!/(n_1 ! times n_2 ! times n_3 ! times ... times n_{12}!)$ where $n_1=6, n_2=4$, and the others are $1!$ (since they're all distinct)?
$endgroup$
– clueless
Jan 19 at 15:52
$begingroup$
And for b, the answer should be $(15!)(6!)$. Is this correct? (assuming the first interpretation is intended)
$endgroup$
– clueless
Jan 19 at 16:01
$begingroup$
@clueless You are correct on both counts.
$endgroup$
– N. F. Taussig
Jan 19 at 16:04
1
$begingroup$
Thank you so much!
$endgroup$
– clueless
Jan 19 at 16:07
add a comment |
$begingroup$
In how many ways can $6$ English books, $4$ science books, $7$ magazines, and $3$ mathematics books be arranged on a shelf if English books are indistinguishable and science books are indistinguishable?
We have a total of $6 + 4 + 7 + 3 = 20$ books. Choose six of the $20$ positions for the English books and four of the remaining $14$ positions for the science books. The remaining ten positions can be filled with books and magazines in $10!$ ways.
$$binom{20}{6}binom{14}{4}10! = frac{20!}{6!14!} cdot binom{14!}{4!10!} cdot 10! = frac{20!}{6!4!}$$
In your attempt, you did not take into account the total number of positions on the shelf.
In how many ways can $6$ English books, $4$ science books, $7$ magazines, and $3$ mathematics books be arranged on a shelf if English books should be together?
If all the books are intended to be distinct (switching the order of the questions would have made this clearer), treat the English books as a single object, so we have $1 + 4 + 7 + 3 = 15$ objects to arrange. Then multiply by the number of ways of arranging the six English books within the block of English books.
If we are still supposed to treat the English books as being indistinguishable and the science books as being indistinguishable, choose six of the $15$ positions for the science books, one of the remaining $8$ positions for the block of English books, then arrange the magazines and mathematics books in the remaining positions.
I believe the first of these two interpretations is intended, but I would have reversed the order of the questions to make that clear.
answered Jan 19 at 13:39
N. F. TaussigN. F. Taussig
44.5k103357
$endgroup$
In how many ways can $6$ English books, $4$ science books, $7$ magazines, and $3$ mathematics books be arranged on a shelf if English books are indistinguishable and science books are indistinguishable?
We have a total of $6 + 4 + 7 + 3 = 20$ books. Choose six of the $20$ positions for the English books and four of the remaining $14$ positions for the science books. The remaining ten positions can be filled with books and magazines in $10!$ ways.
$$binom{20}{6}binom{14}{4}10! = frac{20!}{6!14!} cdot binom{14!}{4!10!} cdot 10! = frac{20!}{6!4!}$$
In your attempt, you did not take into account the total number of positions on the shelf.
In how many ways can $6$ English books, $4$ science books, $7$ magazines, and $3$ mathematics books be arranged on a shelf if English books should be together?
If all the books are intended to be distinct (switching the order of the questions would have made this clearer), treat the English books as a single object, so we have $1 + 4 + 7 + 3 = 15$ objects to arrange. Then multiply by the number of ways of arranging the six English books within the block of English books.
If we are still supposed to treat the English books as being indistinguishable and the science books as being indistinguishable, choose six of the $15$ positions for the science books, one of the remaining $8$ positions for the block of English books, then arrange the magazines and mathematics books in the remaining positions.
I believe the first of these two interpretations is intended, but I would have reversed the order of the questions to make that clear.
answered Jan 19 at 13:39
N. F. TaussigN. F. Taussig
44.5k103357
answered Jan 19 at 13:39
N. F. TaussigN. F. Taussig
44.5k103357
answered Jan 19 at 13:39
N. F. TaussigN. F. Taussig
44.5k103357
answered Jan 19 at 13:39
N. F. TaussigN. F. Taussig
44.5k103357
44.5k103357
$begingroup$
Oh right. But for a, is it also the same as using the formula $n!/(n_1 ! times n_2 ! times n_3 ! times ... times n_{12}!)$ where $n_1=6, n_2=4$, and the others are $1!$ (since they're all distinct)?
$endgroup$
– clueless
Jan 19 at 15:52
$begingroup$
And for b, the answer should be $(15!)(6!)$. Is this correct? (assuming the first interpretation is intended)
$endgroup$
– clueless
Jan 19 at 16:01
$begingroup$
@clueless You are correct on both counts.
$endgroup$
– N. F. Taussig
Jan 19 at 16:04
1
$begingroup$
Thank you so much!
$endgroup$
– clueless
Jan 19 at 16:07
add a comment |
$begingroup$
Oh right. But for a, is it also the same as using the formula $n!/(n_1 ! times n_2 ! times n_3 ! times ... times n_{12}!)$ where $n_1=6, n_2=4$, and the others are $1!$ (since they're all distinct)?
$endgroup$
– clueless
Jan 19 at 15:52
$begingroup$
And for b, the answer should be $(15!)(6!)$. Is this correct? (assuming the first interpretation is intended)
$endgroup$
– clueless
Jan 19 at 16:01
$begingroup$
@clueless You are correct on both counts.
$endgroup$
– N. F. Taussig
Jan 19 at 16:04
1
$begingroup$
Thank you so much!
$endgroup$
– clueless
Jan 19 at 16:07
$begingroup$
Oh right. But for a, is it also the same as using the formula $n!/(n_1 ! times n_2 ! times n_3 ! times ... times n_{12}!)$ where $n_1=6, n_2=4$, and the others are $1!$ (since they're all distinct)?
$endgroup$
– clueless
Jan 19 at 15:52
$begingroup$
Oh right. But for a, is it also the same as using the formula $n!/(n_1 ! times n_2 ! times n_3 ! times ... times n_{12}!)$ where $n_1=6, n_2=4$, and the others are $1!$ (since they're all distinct)?
$endgroup$
– clueless
Jan 19 at 15:52
$begingroup$
And for b, the answer should be $(15!)(6!)$. Is this correct? (assuming the first interpretation is intended)
$endgroup$
– clueless
Jan 19 at 16:01
$begingroup$
And for b, the answer should be $(15!)(6!)$. Is this correct? (assuming the first interpretation is intended)
$endgroup$
– clueless
Jan 19 at 16:01
$begingroup$
@clueless You are correct on both counts.
$endgroup$
– N. F. Taussig
Jan 19 at 16:04
$begingroup$
@clueless You are correct on both counts.
$endgroup$
– N. F. Taussig
Jan 19 at 16:04
1
1
$begingroup$
Thank you so much!
$endgroup$
– clueless
Jan 19 at 16:07
$begingroup$
Thank you so much!
$endgroup$
– clueless
Jan 19 at 16:07
add a comment |
$begingroup$
The number of different permutations of $n$ objects, where $n_{1}$ are of one kind, $n_{2}$ are a different kind $dots$ and there are $k$ different kinds is:
$$
frac{n!}{n_{1}! times n_{2}! times dots times n_{k}!}
$$
answered Jan 19 at 13:09
C_RichmondC_Richmond
863
$endgroup$
add a comment |
$begingroup$
The number of different permutations of $n$ objects, where $n_{1}$ are of one kind, $n_{2}$ are a different kind $dots$ and there are $k$ different kinds is:
$$
frac{n!}{n_{1}! times n_{2}! times dots times n_{k}!}
$$
answered Jan 19 at 13:09
C_RichmondC_Richmond
863
$endgroup$
add a comment |
$begingroup$
The number of different permutations of $n$ objects, where $n_{1}$ are of one kind, $n_{2}$ are a different kind $dots$ and there are $k$ different kinds is:
$$
frac{n!}{n_{1}! times n_{2}! times dots times n_{k}!}
$$
answered Jan 19 at 13:09
C_RichmondC_Richmond
863
$endgroup$
The number of different permutations of $n$ objects, where $n_{1}$ are of one kind, $n_{2}$ are a different kind $dots$ and there are $k$ different kinds is:
$$
frac{n!}{n_{1}! times n_{2}! times dots times n_{k}!}
$$
answered Jan 19 at 13:09
C_RichmondC_Richmond
863
answered Jan 19 at 13:09
C_RichmondC_Richmond
863
answered Jan 19 at 13:09
C_RichmondC_Richmond
863
answered Jan 19 at 13:09
C_RichmondC_Richmond
863
863
add a comment |
add a comment |
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$begingroup$
Have you seen similar questions on the site? By similar, I refer to counting the arrangement of objects of different types. Look at them, and reference any that you find similar in your post above.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:24
1
$begingroup$
Please read the related questions at right, then make an attempt at solving the problem yourself.
$endgroup$
– N. F. Taussig
Jan 19 at 10:45