Logic flaw or not

Question

Claim:

Suppose that you have n books that you want to read in n days, reading one book each day. There are (n!)^2 many ways to do this.

Proof.

Start by selecting your first book; there are n ways in which you can do this. Now, select a day to read that book on: there are also n ways in which you can do this. Assign that book to that day, and remove that book/day from our set of choices. There are n^ 2 many ways in which we can do this; we have n books and n days, and any possible pairing is valid. Now, do the same thing again. There are n − 1 choices for our second book, as we have removed one choice already; there are also n − 1 choices for the day on which we read this book. There are (n − 1)^2 many ways in which we can do this, by the same logic as above. Repeat this process! In total, we have n 2 · (n − 1)2 · . . . · 2 2 · 1 2 = (n!)^2 many ways in which to read our books.

Struggling to tell if there is a flaw in the logic or not. My example: Suppose we have 2 books, that I want to read in 2 days... one for each day... there are supposedly 2^2 ways to do this...

Book A - Monday

Book B - Monday

Book A -Tuesday

Book B Tuesday

Select a book 2 ways to do this, 2 ways to select a day. So assigned Book A to Monday , and removed it from set. There is one choice for book and one choice for day, only one way to do this now... Does this seem right or am i missing something would greatly appreciate feedback :)

Answer 1

You overcount.

You say that Book A on Monday is a different way to read the books from reading Book B on Tuesday. But clearly those are actually the same: you read Book A and Monday if and only if you read Book B on Tuesday.

Answer 2

It's $n!$, not $n!^2$. An explanation of how you're overcounting things:

Say there are two books, $A$ and $B$. There are clearly only two possibilities, $AB$ and $BA$. Your overcounting arises because you're making an irrelevant distinction regarding which is the "first book" - this is irrelevant because, curiously, whatever you mean by saying a certain book is the "first book", you're not requiring that the first book be read first.

Say $b_1$ is the one you call "first" and $b_2$ is the other. The four cases you're counting are these:

(i) $b_1=A, b_2=B$, read in order $b_1b_2$.

(ii) $b_1=A, b_2=B$, read in order $b_2b_1$.

(iii) $b_1=B, b_2=A$, read in order $b_1b_2$.

(iv) $b_1=B, b_2=A$, read in order $b_2b_1$.

Buut in fact in terms of the actual question, (i) and (iv) are the same, since they both say $AB$, and similarly (ii) and (iii) both say $BA$.

Moral: Giving one of the books the meaningless designation "first book". meaningless since the "first book" is not the one that's read first, was a very bad idea, likely to lead to confusion. In fact you used this technique to confuse yourself...

Answer 3

For the first day, you have to read one book among $n$ books. So this job can be done in $n$ ways.

Remove the book that you have read in the first day.

For the second day, you have to read one book among $n-1$ books. So this job can be done in $n-1$ ways.

Again remove the book that you have read in the second day.

Thus, at the $n^{th}$ day, you have only one book to read. So this job can be done in $1$ way.

Hence we conclude that the total number of ways $=n(n-1)(n-2)... 2.1=n! $.