Brisa Lyons

2023-01-01

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

Belinda Solomon

Beginner2023-01-02Added 11 answers

Explanation: Assuming the books can be separated:

The leftmost book has five options.

For each of these choices there are 4 choices for the second book from the left;

giving$5\times 4=20$ 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$5\times 4\times 3=60$ 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$5\times 4\times 3\times 2=120$ choices for the first four books on the left.

This only leaves 1 choice (not really a "choice") for the fifth book;

giving$5\times 4\times 3\times 2\times 1=120$ choices for arranging the books (from left to right).

The leftmost book has five options.

For each of these choices there are 4 choices for the second book from the left;

giving

Having chosen the first 2 books there are 3 choices for the third book from the left;

giving

Having chosen the first 3 books there are 2 choices for the fourth book from the left;

giving

This only leaves 1 choice (not really a "choice") for the fifth book;

giving

Isaias Black

Beginner2023-01-03Added 1 answers

Assume that the books can be told apart. 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:

$n(n-1)(n-2)\cdots \left(3\right)\left(2\right)\left(1\right)$

$=n!$

With five distinct books (you can still distinguish one book from another by looking at it), we have the following:

$5!=1\cdot 2\cdot 3\cdot 4\cdot 5=\mathbf{120}$ configurations

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.

With five distinct books (you can still distinguish one book from another by looking at it), we have the following:

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.

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