Extracting numbered balls from box Problem
A box contains $N$ balls numbered from $1$ to $N$. Let $1 \leq n  \leq N$ and suppose $n$ balls are extracted successively without reposition. Calculate the probability that the numbers of the extracted balls in the order they were extracted form a strictly increasing sequence. Then calculate the probability of this event in case the extraction is with reposition
I am a bit lost with this problem. In the first case, where the extraction is without repositon, I have to pick $n$ balls from $N$ and there is only one way to order these $n$ balls given the condition of the problem. So if $A$ is the event, then $$P(A)=\binom{N}{n}$$
I don't know what to do in the other case, I would appreciate suggestions and if someone could tell me whether I've calculated the probability correctly in the first case. Thanks in advance.
 A: You’ve correctly calculated the number of successful outcomes for the first case, but that’s not the probability: for the probability you need to divide $\binom{N}n$ by the number of possible outcomes, which is $\binom{N}nn!$, since each set of $n$ balls can be drawn in $n!$ different orders. Thus, the actual probability is $\frac1{n!}$.
You can also find this by following a different line of reasoning: no matter which $n$ balls you draw, there is exactly one order of drawing them that results in an increasing sequence, so the probability of getting it is $\frac1{n!}$, independent of which $n$ balls are drawn.
For the case of drawing with replacement you shouldn’t have much trouble counting the total number of possible outcomes, so I’ll leave that part to you. To calculate the number of successful outcomes you can use what you’ve already done: a successful outcome must involve drawing $n$ different balls, and they must be drawn in exactly the right order. Thus, there is one successful outcome for each possible set of $n$ different balls. Can you take it from there?
