Question

In how many different ways can 5 books be arranged on a shelf?

Answer 1

`120`

Explanation:

Assuming that the books are distinguishable:

There are `5` choices for the left-most book.

For each of these choices there are `4` choices for the second book from the left;
giving `5xx4 =20` choices for the first two books on the left.

Having chosen the first 2 books there are `3` choices for the third book from the left;
giving `5xx4xx3=60` choices for the first three books on the left.

Having chosen the first 3 books there are `2` choices for the fourth book from the left;
giving `5xx4xx3xx2=120` choices for the first four books on the left.

This only leaves `1` choice (not really a "choice") for the fifth book;
giving `5xx4xx3xx2xx1=120` choices for arranging the books (from left to right).

Answer 2

Suppose the books are distinguishable. Once you place one on the shelf, there are `n-1` remaining books to place, where `n` is the starting number of books. Once you place a second on the shelf, there are `n-2` remaining books to place. Repeat until `n - k = 1`, and you have the definition of a factorial:

`n(n-1)(n-2)cdots(3)(2)(1)`

`= n!`

Therefore, with five distinguishable books (interchanging them, you can still tell one book from another by inspection), we have:

`5! = 1*2*3*4*5 = bb120` configurations

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If, for some reason, you have blurry vision and the books are indistinguishable, we have to account for redundant configurations after assuming distinguishability.

If we place one book in the shelf, it can also occupy the space of the other `n-1` spots on the shelf, giving us `n` arrangements of the one book. Thus, we divide by `n` to account for the identical arrangements.

Then, for book `2`, there are `n-1` identical arrangements, and so on, all the way down to the last book.

Thus, for `n!` otherwise distinguishable arrangements, we have to divide by `n!` to account for redundant configurations, and we simply have one arrangement of five indistinguishable books possible.