Probability to draw a group of balls which contains a specific ball 
A bin holds n balls, labeled with the numbers $1, 2, . . . , n$. Exactly $m$ balls are being
  sampled uniformly at random from the bin. Let $M$ be the maximum number that was drawn.
  
  
*
  
*Compute the distribution of $M$, when the samples are being made without replacement.
  

My answer:
We need to calculate $P(M=k)$
I have 2 answers which result in 2 different values.


*

*We must draw the $k$'th ball ($1$ way to do it) and then we need to choose the rest of the balls ($\binom{k-1}{m-1}$). There are $\binom{n}{m}$ ways to choose $m$ balls from $n$ balls:


$$\frac{1\cdot\binom{k-1}{m-1}}{\binom{n}{m}}$$


*We must draw the $k$'th ball $\frac{1}{n}$  and then we need to choose the rest of the balls $\binom{k-1}{m-1}$. The probability to choose each of these balls is:


$$ \frac{1}{n-1}\cdot\frac{1}{n-2}\cdots\frac{1}{n-m} $$
So the final result is:
$$ \frac{1}{n}\cdot\binom{k-1}{m-1}\cdot\frac{1}{n-1}\cdot\frac{1}{n-2}\cdots\frac{1}{n-m} = $$
$$ \frac{1}{n}\cdot\binom{k-1}{m-1}\cdot\frac{(n-m-1)!}{n!} $$

My question - Each result is different. Which is the correct one and why is the other not correct?
 A: The first method is the correct one, but method number 2 can be fixed. I will try to point out the errors and fix them:

  
*
  
*The product $$\frac1{n-1}\cdot\frac1{n-2}\cdots\frac1{n-m}$$ should have stopped at $n-m+1$ instead, since otherwise you would draw $m+1$ balls as a total.
  
*The fraction $1/n$ should be included as part of $n!$ in the denominator in the last line, since the other fractions only have $n-1,n-2$ etc.
  
*Finally you should multiply the whole thing by the $m!$ different orders in which those $m$ balls could be drawn.

After doing this, the two actually agree since they both become:
$$
\binom{k-1}{m-1}\cdot\frac{(n-m)!\cdot m!}{n!}
$$
if you look carefully.

Note that method 1 considers combinations where order does not matter, whereas method 2 thinks through probabilities for each specific sequence of events and hence permutations where order does matter. That is what prompted the correction of a factor $m!$ in the method 2 approach. In method 1 that factor $m!$ is just already an element of the formula for the orderless $\binom{n}{m}$.
A: I would say $$P(M = k)=P(M \le k)-P(M \le k-1) =\dfrac{k \choose m}{n \choose m}-\dfrac{k-1 \choose m}{n \choose m} = \dfrac{m{k \choose m}}{k{n \choose m}}=\dfrac{{k-1 \choose m-1}}{{n \choose m}}$$ so the first option in the question
The second option is too small and does not add up to $1$
