Trouble understanding on how to calculate all possible events of sample space S While doing some exercises I had following question:
How many possible events can be defined if the sample space S looks as follows:


*

*$S = \{1,2,3\}  $

*$S = \{1,2,3,4\}  $

*$S = \{s_1, ..., s_n\}$
The solution is $8$, $16$, $2^n$.
Can someone explain me why the solution is $2^n$?
 A: I think it’s because each number in the sample space is either part of the event or not, which is two possibilities. So with the three numbers 1, 2 and 3, there are 2 x 2 x 2 possibilities which is 2$^3$ This is then generalized with n numbers to 2$^n$
A: Hint: You can count the events. The sample space $S$ contains $n$ possible outcomes. What is the event space at the first case? 
It is $\{\emptyset,1,2,3,12,13,23,123\}$. $\emptyset$ is the empty set (Never forget it!). It constains $2^3=8$ events. If you say that $k$ is the number of elements of a sequence then it can be written as a sum: $\sum\limits_{k=0}^n \binom{n}{k}=2^n$. 
A: If the set has $n$ elements, then you are looking for the number $2^n$. 
If you know some combinatorics, it follows from: 
Let $S$ the sample space with $n$ elements 
total events: all individual events + all combinations of 2 events + all combinatinations of 3 events + .. + all combination of $n-1$ events + all combinations of $n$ events (only 1 comb.)


*

*all individual events: this is the combinatoric of $n$ in $1$

*all combinations of 2 events: this is the combinatoric of $n$ in $2$ 
-all combinations of 3 events: this is the combinatoric of $n$ in $3$ 

*...


So you need the number :
$${{n}\choose{0}}+{n\choose{1}}+{n\choose{2}}+{n\choose{3}}+\cdots+{n\choose{n-1}}+{n\choose{n}}=2^n$$
