# Simple probability proof

1. Oct 2, 2013

### phospho

An enclyclopededia consisting of n ($n \geq 4$) similar volumes is kept on a shelf with the volumes in correct numerical order: that is, with volume 1o n the left, volume 2 next, and so on. The volumes are all taken down for cleaning and are replaced on the shelf in a random order. Prove that the probabilities of finding exactly n, (n-1), (n-2), (n-3) volumes in their correct positions on the shelf are, respectively, $\dfrac{1}{n!}$, $0$, $\dfrac{1}{(n-2)!2}$, $\dfrac{1}{(n-3)!3}$

I've done all of this, except I cannot construct a proof for (n-1) and 0

It seems very trivial, if one of the books is in the wrong place, then that means another book is in the wrong place, which means another book is in the wrong place and so on, therefore the probability of a book being in the correct position is 0, but how do I go about constructing an actual proof for it?

2. Oct 2, 2013

### LCKurtz

Yes.

Why? Couldn't just two books be swapped?

3. Oct 2, 2013

### phospho

I guess, but that still leaves the probability to be 0, as only 1 book being in the wrong place means at least two are in the incorrect position

4. Oct 2, 2013

### LCKurtz

That's correct. You can't have just one in the wrong place, and that's why that probability is zero.

5. Oct 2, 2013

### phospho

Is that a proof?

6. Oct 2, 2013

### LCKurtz

It is certainly the essence of a proof. You could flesh it out something like: Suppose a book, call it book $p$, is in the wrong place, say it is in position $q\ne p$. Then book $q$ is not in its correct place, so...

7. Oct 3, 2013

### haruspex

Seems unnecessarily complicated. If n-1 books are in their correct positions, the remaining position is not the correct one for any of them, so it must the correct position for the remaining volume.