Combinatorial Distribution with random sample I believe I have no idea what kind of distribution this is and that is what I would like.
Balls are numbered 1 to $N$. We select a sample of $n$ at random. Let $Y$ be the largest number in the sample. 
Find the distribution of $Y$. What is this distribution in the first place? 
Show $$\binom{N}{n+1}=\sum^{N-1}_{k=n} \binom{k}{n}.$$ So this is what I have so far. 
$${N \choose {n+1}}=\frac{N!}{(n+1)! (N-n-1)!}=\frac{N(N-1)(N-2)\cdots(N-n)}{(n+1)n!}.$$
And
\begin{align}
& \sum^{N-1}_{k=n} {k\choose n} = {n \choose n}+{{n+1}\choose n}+{{n+2}\choose n}+\cdots+{{N-1}\choose n} \\[8pt]
= {} & \frac{n!}{n!(n-n)!} + \frac{(n+1)!}{n!1!} + \frac{(n+2)!}{2!} + \frac{(N-1)!}{n!(N-1-n)!}.
\end{align}
I was able to factor out a $\frac{1}{n!}$
but other than that I am stuck.
The second part to that is conclude that when you add $(N \choose n)$ you get $\Sigma^{N}_{k=n}$$k\choose n$. But this is fairly simple.
Lastly, write down how to solve $E(Y)$. Use the second part in above to find a formula for the expected value of Y. se formula to find expected value for largest umber if sample 30 out of 100.
This is were I am also confused because I do not know of the distribution. I think it is a simple random sample, but I am unfamiliar with how to work with those. 
 A: Suppose $n$-subsets of $\{1, 2, \ldots, N\}$, with $N \ge n$, are selected at random (without replacement).  Let $Y$ be the largest value in the selected $n$-subset.  The sample space of elementary outcomes consists of $$\binom{N}{n}$$ $n$-subsets; this is easy to see.  Now consider if $Y = y$ for some $n \le y \le N$.  How many such $n$-subsets are there for which the largest element is $y$?  If we fix this maximum value, there are $n-1$ remaining elements to be selected from the set $\{1, 2, \ldots, y-1\}$, hence there are $$\binom{y-1}{n-1}$$ such subsets with maximal value $y$.  Therefore, $$\Pr[Y = y] = \frac{\binom{y-1}{n-1}}{\binom{N}{n}}, \quad n \le y \le N$$ and $0$ otherwise.  This is the desired probability mass function.
To calculate the expectation of $Y$, we write $$\begin{align*}\operatorname{E}[Y] &= \sum_{y=n}^N y \Pr[Y = y] = \sum_{y=n}^N y \cdot \frac{\binom{y-1}{n-1}}{\binom{N}{n}} \\ &= \frac{1}{\binom{N}{n}} \sum_{y=n}^N y \binom{y-1}{n-1} = \frac{1}{\binom{N}{n}} \sum_{y=n}^N y \cdot \frac{(y-1)!}{(n-1)! (y-n)!} \\ &= \frac{1}{\binom{N}{n}} \sum_{y=n}^N n \cdot \frac{y!}{n!(y-n)!} = \frac{n}{\binom{N}{n}} \sum_{y=n}^N \binom{y}{n} \\ &= \tag{1} \frac{n}{\binom{N}{n}} \sum_{y=n+1}^{N+1} \binom{y-1}{(n+1)-1}. \end{align*}$$  All I have done in the last step is to increment the index of summation by $1$, and compensate by decrementing the index in the summand by $1$:  this is analogous to writing, for example, $$\sum_{y=0}^5 y^2 = \sum_{y=1}^6 (y-1)^2.$$  The reason to do this is to exploit the fact that we already know $$\sum_{y=n}^N \Pr[Y = y] = \sum_{y=n}^N \frac{\binom{y-1}{n-1}}{\binom{N}{n}} = 1,$$ because the sum of the probabilities of all the possible outcomes of $Y$ must be $1$.  Therefore, multiplying both sides by $\binom{N}{n}$ gives the identity $$\tag{2} \sum_{y=n}^N \binom{y-1}{n-1} = \binom{N}{n}.$$  But here, the summand has the form $\binom{y-1}{n-1}$, not $\binom{y}{n}$ as we had in the penultimate step in our calculation of the expectation (just before we wrote $(1)$).  So we "massage" the summand into the desired form by rewriting it, and shifting the index so as not to change the value of the sum.  Then we recognize that $$\sum_{y=n+1}^{N+1} \binom{y-1}{(n+1)-1}$$ corresponds to the identity $(2)$ we want to use, except with the substitution $$n \to n+1, \quad N \to N+1,$$ hence the desired sum has value $$\sum_{y=n+1}^{N+1} \binom{y-1}{(n+1)-1} = \binom{N+1}{n+1}.$$  Therefore, continuing from $(1)$, $$\operatorname{E}[Y] = \frac{n}{\binom{N}{n}} \binom{N+1}{n+1} = n \cdot \frac{(N+1)!}{(n+1)!(N-n)!} \cdot \frac{n! (N-n)!}{N!} = \frac{n(N+1)}{n+1}.$$
A: For the distribution of $Y$, we have $$\operatorname{Pr}(Y \leq k) = {k\choose n}/{N\choose n}.$$ This is because if $Y\leq k$, we are choosing $n$ balls from the balls numbered from 1 to $k$, and there are $k\choose n$ many ways to do that.
A: The first part is trivial: If we want to choose $n+1$ numbers from $1,\dots,N$, we can $\binom{N}{n+1}$ combinations. On the other hand, we can first fix the largest number $M$ we can choose, that $M$ can be one of $n+1,n+2,\dots,N$, in each case, we can have $\binom{M-1}{n}$. In other words, $\binom{N}{n+1}=\sum_{k=n}^{N-1}\binom{k}{n}$.
The expectation $\mathbb{E}(Y)=\dfrac{\binom{n}{n}}{\binom{N}{n+1}}(n+1)+\dfrac{\binom{n+1}{n}}{\binom{N}{n+1}}(n+2)+\dots+\dfrac{\binom{N-1}{n}}{\binom{N}{n+1}}(N)=\dfrac{\binom{n}{n}(n+1)+\binom{n+1}{n}(n+2)+\dots+\binom{N-1}{n}(N)}{\binom{N}{n+1}}=\dfrac{(n+1)\left(\binom{n+1}{n+1}+\binom{n+2}{n+1}+\dots+\binom{N}{n+1}\right)}{\binom{N}{n+1}}=\dfrac{(n+1)\binom{N+1}{n+2}}{\binom{N}{n+1}}=\dfrac{(n+1)(N+1)}{n+2}.$
