In how many way we can arrange 4 books on 7 different shelves - with no specific demandings?

Question

I was given an unconventional combinatorics question I was struggling with. Suppose we have 4 books and 7 different bookshelves. In how many ways the books can be arranged without loss of generality?

I can either put all 4 of them on 1 shelf, or either split each one in a different shelf. In a case such as that, how do I consider the shelves that can't be picked? how to sum the number of options I have left when I do pick $k < 4$ books on 1 shelf (which are arranged in $k!$ options) and left with 6 different shelves and $4-k$ books?

Best regards for the helpers.

Answer 1

(Assume all books and shelves identical)

So you can have $(4,0,...,0)$ - 7 ways to choose which shelf this is

$(3,1,0,...,0)$ - 42 ways to choose the two shelves i.e. 6*7

$(2,2,0,...,0)$ - 7C2 ways to choose the two shelves (i.e. 6*7/2)

$(2,1,1,0,...,0)$ - 7*(6C2)

$(1,1,1,1,0,...,0)$ - 7C4

Now add all of these solutions together to get the final answer.

Answer 2

Each book can be put into $7$ different book shelves. So the total number of ways is $7^4.$

Answer 3

In the following, I will assume that the books as well as the shelves are distinguishable.

The four books may be arranged in $4!$ ways. We must choose how many books go on each of the seven shelves. Let $x_j$, $1 \leq j \leq 7$, be the number of books placed on the $j$th shelf. Then $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 4$$ is an equation in the nonnegative integers. A particular solution of the equation corresponds to the placement of six addition signs in a row of four ones. For instance, $$+ + 1 1 + + + 1 + 1$$ corresponds to the solution $x_1 = 0$, $x_2 = 0$, $x_3 = 2$, $x_4 = 0$, $x_5 = 0$, $x_6 = 1$, $x_7 = 1$, while $$1 + + 1 + + 1 + + 1$$ corresponds to the solution $x_1 = 1$, $x_2 = 0$, $x_3 = 1$, $x_4 = 0$, $x_5 = 1$, $x_6 = 0$, $x_7 = 1$. The number of such solutions is the number of ways we can place six addition signs in a row of four ones, which is $$\binom{4 + 7 - 1}{7 - 1} = \binom{10}{6}$$ since we must select which six of the ten places required for four ones and six addition signs will be filled with addition signs.

Thus, the number of ways of placing four distinct books on seven shelves is $$4!\binom{10}{6}$$

Alternatively, we must arrange four distinct books and six identical dividers in a row - the dividers separate the books onto different shelves. We can select the positions of the books in $\binom{10}{4}$ ways. The books can be arranged in the selected positions in $4!$ ways. The dividers must fill each of the remaining six positions. Hence, there are $$\binom{10}{4}4!$$ ways to arrange four distinct books on seven shelves.