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