Q33. If n apples, no two of the same weight, are lined up at random on a table, what is the probability that they are lined up in order of increasing weight from left to right?

Answer: 1/n!

May I ask what is the method to derive the expression? At first I was tricked and chose (1/n)^n.

Is this reasoning valid:

First put the first apple.

A1

The next apple can have two choices, left of it, or right of it (with gaps in between), so probability is 1/2.

A1 A2

Now, the next apple (A3) has three choices, left of A1, middle, or right of A2, one of which is correct, so prob is 1/3

...

So on, until last apple (An), which has n choices, so prob is 1/n.

Since the probability is independent (why?), so multiplying the probabilities, we get 1/n!

Thanks.

## 9367 Q33

### Re: 9367 Q33

Just find the total number of ways you can position the apples in a row.

Look at the first position. You have n choices of apples.

Now look at the second position. You will now only have n-1 choices since u have already placed an apple.

Continue doing this and you get n(n-1)(n-2)....(2)(1) = n! ways of placing the apples.

Now what you need is one specific way out of the n! ways of placing apples.

So the probability is thus 1/n! .

Look at the first position. You have n choices of apples.

Now look at the second position. You will now only have n-1 choices since u have already placed an apple.

Continue doing this and you get n(n-1)(n-2)....(2)(1) = n! ways of placing the apples.

Now what you need is one specific way out of the n! ways of placing apples.

So the probability is thus 1/n! .