How many ways to arrange four types of books on the shelf so that books of the same type are together?
Janet has $10$ different books that she is going to put on her bookshelf. Of these, $4$ are Book C, $3$ are Book B, $2$ are Book S and $1$ Book P. Janet 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?
My workings are
$4! cdot 3! cdot 2! cdot 1!$
but the actual answer has an additional multiplication by $4!$
Why is this so? My guess is because we need the $4$ books to come together, let's say for book C. But isn't that "shown" by the working of $4!$ for Book C and $3!$ for Book B and so on?
combinatorics permutations
edited Nov 24 at 11:10
N. F. Taussig
43.5k93355
asked Nov 24 at 8:16
mutu mumu
374
add a comment |
Janet has $10$ different books that she is going to put on her bookshelf. Of these, $4$ are Book C, $3$ are Book B, $2$ are Book S and $1$ Book P. Janet 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?
My workings are
$4! cdot 3! cdot 2! cdot 1!$
but the actual answer has an additional multiplication by $4!$
Why is this so? My guess is because we need the $4$ books to come together, let's say for book C. But isn't that "shown" by the working of $4!$ for Book C and $3!$ for Book B and so on?
combinatorics permutations
edited Nov 24 at 11:10
N. F. Taussig
43.5k93355
asked Nov 24 at 8:16
mutu mumu
374
Problems of enumeration should be tagged combinatorics rather than probability.
– N. F. Taussig
Nov 24 at 11:10
add a comment |
Janet has $10$ different books that she is going to put on her bookshelf. Of these, $4$ are Book C, $3$ are Book B, $2$ are Book S and $1$ Book P. Janet 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?
My workings are
$4! cdot 3! cdot 2! cdot 1!$
but the actual answer has an additional multiplication by $4!$
Why is this so? My guess is because we need the $4$ books to come together, let's say for book C. But isn't that "shown" by the working of $4!$ for Book C and $3!$ for Book B and so on?
combinatorics permutations
edited Nov 24 at 11:10
N. F. Taussig
43.5k93355
asked Nov 24 at 8:16
mutu mumu
374
Janet has $10$ different books that she is going to put on her bookshelf. Of these, $4$ are Book C, $3$ are Book B, $2$ are Book S and $1$ Book P. Janet 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?
My workings are
$4! cdot 3! cdot 2! cdot 1!$
but the actual answer has an additional multiplication by $4!$
Why is this so? My guess is because we need the $4$ books to come together, let's say for book C. But isn't that "shown" by the working of $4!$ for Book C and $3!$ for Book B and so on?
combinatorics permutations
combinatorics permutations
edited Nov 24 at 11:10
N. F. Taussig
43.5k93355
asked Nov 24 at 8:16
mutu mumu
374
edited Nov 24 at 11:10
N. F. Taussig
43.5k93355
asked Nov 24 at 8:16
mutu mumu
374
edited Nov 24 at 11:10
N. F. Taussig
43.5k93355
edited Nov 24 at 11:10
N. F. Taussig
43.5k93355
edited Nov 24 at 11:10
N. F. Taussig
43.5k93355
43.5k93355
asked Nov 24 at 8:16
mutu mumu
374
asked Nov 24 at 8:16
mutu mumu
374
asked Nov 24 at 8:16
mutu mumu
374
374
Problems of enumeration should be tagged combinatorics rather than probability.
– N. F. Taussig
Nov 24 at 11:10
add a comment |
Problems of enumeration should be tagged combinatorics rather than probability.
– N. F. Taussig
Nov 24 at 11:10
Problems of enumeration should be tagged combinatorics rather than probability.
– N. F. Taussig
Nov 24 at 11:10
Problems of enumeration should be tagged combinatorics rather than probability.
– N. F. Taussig
Nov 24 at 11:10
add a comment |
1 Answer
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You can visualise it this way: there need to be four blocks and each block contains books of the same type.
One possible arrangement might be like this:
$$C_1 C_2 C_3 C_4 | B_1 B_2 B_3 | S_1 S_2 | P_1$$
Now note that I had no particular reason to put the C block first, followed by B,S and P.
The books within each block can be arranged in $4!3!2!1!$ ways AND the block themselves can be arranged in $4!$ ways. That's where your extra $4!$ is coming from.
answered Nov 24 at 8:28
Gokul
332318
add a comment |
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You can visualise it this way: there need to be four blocks and each block contains books of the same type.
One possible arrangement might be like this:
$$C_1 C_2 C_3 C_4 | B_1 B_2 B_3 | S_1 S_2 | P_1$$
Now note that I had no particular reason to put the C block first, followed by B,S and P.
The books within each block can be arranged in $4!3!2!1!$ ways AND the block themselves can be arranged in $4!$ ways. That's where your extra $4!$ is coming from.
answered Nov 24 at 8:28
Gokul
332318
add a comment |
You can visualise it this way: there need to be four blocks and each block contains books of the same type.
One possible arrangement might be like this:
$$C_1 C_2 C_3 C_4 | B_1 B_2 B_3 | S_1 S_2 | P_1$$
Now note that I had no particular reason to put the C block first, followed by B,S and P.
The books within each block can be arranged in $4!3!2!1!$ ways AND the block themselves can be arranged in $4!$ ways. That's where your extra $4!$ is coming from.
answered Nov 24 at 8:28
Gokul
332318
add a comment |
You can visualise it this way: there need to be four blocks and each block contains books of the same type.
One possible arrangement might be like this:
$$C_1 C_2 C_3 C_4 | B_1 B_2 B_3 | S_1 S_2 | P_1$$
Now note that I had no particular reason to put the C block first, followed by B,S and P.
The books within each block can be arranged in $4!3!2!1!$ ways AND the block themselves can be arranged in $4!$ ways. That's where your extra $4!$ is coming from.
answered Nov 24 at 8:28
Gokul
332318
You can visualise it this way: there need to be four blocks and each block contains books of the same type.
One possible arrangement might be like this:
$$C_1 C_2 C_3 C_4 | B_1 B_2 B_3 | S_1 S_2 | P_1$$
Now note that I had no particular reason to put the C block first, followed by B,S and P.
The books within each block can be arranged in $4!3!2!1!$ ways AND the block themselves can be arranged in $4!$ ways. That's where your extra $4!$ is coming from.
answered Nov 24 at 8:28
Gokul
332318
answered Nov 24 at 8:28
Gokul
332318
answered Nov 24 at 8:28
Gokul
332318
answered Nov 24 at 8:28
Gokul
332318
332318
add a comment |
add a comment |
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Problems of enumeration should be tagged combinatorics rather than probability.
– N. F. Taussig
Nov 24 at 11:10