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In how many ways can $n$ distinct books be arranged on $k$ distinct bookshelves if at least one shelf is to be empty?

Well, I tried writing out the possible number of ways for the cases that $n=1$ and $n=2$, for various $k$ values, and it looks like the answer is ${k \choose n} + k$. Is this correct? And also, what other ways are there to solve a problem like this?

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Solve the problem:

  • for any arrangement of books
  • for the case when no shelf is empty

then subtract the second result from the first.


To solve the first part, represent the arrangements by strings of the form $$\text{ABC*DEFG*HIJK}\ldots$$ where $*$ represents the transition from one shelf to the shelf below.

To solve the second part, imagine that you have ordered the books in one of $n!$ ways, then choose $k-1$ of the $n-1$ apostrophes below, and replace them with asterisks: $$\mathrm{A'B'C'D'E'F'}\ldots\mathrm{'Z}$$

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