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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?

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  • $\begingroup$ Problems of enumeration should be tagged combinatorics rather than probability. $\endgroup$ Nov 24 '18 at 11:10
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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.

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