Consider:
$E_j$: the maximum of the values obtained on $k + 1$ draws is $j$, $j = 1,\ldots,n$,
$F_j$: all the draws are done on $\{1,\ldots,j\}$, $j = 1,\ldots,n$,
$F_0 = \emptyset$.
Calculate $|S|$.
Here, I think that if we execute the action steps by steps, we'll have n possibilities each time so n^(k+1) possibilities after (k+1) times drawing balls. Then $$ |S| = n^{k+1} $$ The thing is the number of balls drawn is not defined, so I think to another situation.
I'll divide the experience in many cases where: draw $1$ ball, $2$ balls then draw $n$ balls. So far, we receive $|S| = 1^{k+1} + 2^{k+1}+\ldots+n^{k+1}$.
But I am not sure about my procedure above.Write $S$ as a set operation on $E_1,\ldots, E_n$.
Like for the first question, I'll solve the problem in many cases where $j=1,2,\ldots,n$.
With $j=1$, we have $|E_1| = k+1$ because when $1$ is the maximum value observed, we draw only $1$ ball. After $k+1$ times, there are $k+1$ possibilities where $E_1$ can be performed. With the same reason, $$ |E_n| = (k+1) 2^{n-1}. $$ Then $$ |E_j| = (k+1) (1+4+8+...+2^{n-1}). $$Write $E_j$ as a set operation on $F_{j-1}; F_j$, $j = 1,\ldots,n$.
- Calculate $|F_j|$, $j = 0,\ldots,n$.
- Calculate $|E_j|$, $j = 1,\ldots,n$.
- Calculate $|S|$, through the computation of $|E_j|$, $j = 1,\ldots,n$ and using the binomial of Newton.
Overall, I don't really understand what the event $F_j$ is. And please correct me and give me some suggestions, please. Ty