How many ways can 5 different types of books be displayed on a bookshelf?

In how many ways can you arrange 8 distinct books in 4 bookshelves? Here the order of arrangement also matters.

Let us first simplify and solve this problem in stages. We will begin by arranging 5 distinct books in 2 bookshelves.

If there was only one bookshelf we can arrange the 5 books in 5! Ways = 120 different ways.

Now let’s arrange 5 books in 2 bookshelves.

One way of doing this is to arrange all the 5 books in the first bookshelf and none in the second. Or we can select any one of the five books and keep it in the first bookshelf and arrange the remaining 4 books in the second bookshelf. Or select any 2 of the 5 books and arrange them in the first bookshelf and arrange the remaining 3 in the second bookshelf.

Let us list out all the possible selections and arrangements:

1) All the 5 books are arranged in the first bookshelf. There will be no books in the second shelf and this will be arranged in 0! = 1 ways.

This can be done in 5! Ways.

2) Out of the 5 books we select any one book and put it in the second shelf. This is the same as selecting 4 books out of 5 and arranging them in the first bookshelf.

We can select 4 books out of 5 in

ways. And these can be arranged in 4! ways in the first bookshelf. The one book in the second bookshelf can be arranged in 1! Ways. Therefore, in this case we have total possible arrangements as

3) Out of the 5 books we now select any 3 books and arrange them in the first bookshelf in 3! ways. The remaining 2 books will be arranged in the second bookshelf in 2! ways. Therefore, in this case we have total possible arrangements as

4) Out of the 5 books we now select any 2 books and arrange them in the first bookshelf in 2! ways. Remaining 3 books will be arranged in the second bookshelf in 3! ways. Therefore, in this case we have total possible arrangements as

5) We now select any one of the 5 books and keep it in the first shelf. Remaining 4 books will be arranged in the second shelf in 4! ways. Therefore, in this case we have total possible arrangements as

6) In the last case we will keep all the 5 books in the second shelf. This can be arranged in the second shelf in 5! Ways.

Therefore, the total number of ways in which we can arrange 5 distinct books in 2 shelves is = 6 x 120 = 720.

Bookshelf I

Bookshelf II

A B C

D E

Figure 1

Let us now look at the problem in a different way. Let us place both the bookshelves besides each other, refer Figure 1. Let A, B, C, D and E represent 5 distinct books. As can be seen in the figure, both the bookshelves share a common partition. Let us represent the partition with letter P. The arrangement in Figure 1 can now be represented as in Figure 2.

A B C P D E

Figure 2

As can be seen from Figure 2, we can now consider partition P as one of the books. Six things can be arranged in 6! different ways. Thus, we can arrange 5 distinct books in 6! = 720 different ways. We got the same answer using two different methods.

Let us use the wisdom from the second method to solve our problem.

Figure 3

Let us place 4 bookshelves as shown in Figure 3. Let us represent the partitions between bookshelves 1 and 2, between 2 and 3 and between 3 and 4 by P1, P2 and P3 respectively. The arrangement in Figure 3 can now be represented as in Figure 4.

A B C D P1 E P2 P3 F G H

Figure 4

11 distinct things shown in Figure 4 can be arranged in 11! different ways. But the 3 partitions P1, P2 and P3 cannot change their order. The 3 partitions can be arranged in 3! ways.

Thus, the answer to our problem is = 11! / 3! = 66,52,800 ways.

In general, if we have b distinct books and n bookshelves, where b > n, then they can be arranged in

[b + (n – 1)]! / (n – 1)! different ways

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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

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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From the problem statement $n = 5$ and $r=2$

When I apply the values in the formula $^nP_r=\frac{n!}{(n-r)!}$

I get the value $\frac{5!}{(5-2)!} = \frac{5*4*3*2*1}{3*2*1} = 20$

And it is a wrong answer. How to solve this problem correctly?