How many combinations of 6 items are possible?

Your are asking the number of subsets of a set with n elements.{1,2,3,...,n} Each subset can be represented by a binary string, e.g for the set {1,2,3,4,5,6} the string 001101 means the subset that

does not contain the element 1 of the set, because the 1st left character of the string is 0

does not contain the element 2 of the set, because the 2nd left character of the string is 0

does contain the element 3 of the set, because the 3rd left character of the string is 1

does contain the element 4 of the set, because the 4th left character of the string is 1

does not contain the element 5 of the set, because the 5th left character of the string is 0

does contain the element 6 of the set, because the 6th left character of the string is 1

so 001101 means the subset {3,4,6}. Therefore there asre as many subsets as strings of length n. With n binary digits one can count from 0 to 2^n-1, therefore there are 2^n such strings and 2^n subsets of {1,....,n}. 00...0 means the empty subset. if you dont want count the empty subset then you have only 2^n-1 subsets.

$2^6$. Think of it like a garden of forking paths where at the first fork you have to decide whether or not to include 1, then 2, then 3... With each choice the number of possibilities doubles.

My thought was:

Each combination of $k$ elements $n$ by $n$ can be express by $^kC_{n}$.As I understood you want to make sets of $1,2,3,4,5$ and $6$ elements from a sample of $6$ elements without reapeting.So there is a line in pascal's triangle where $k=6$ and $n \in \{ 1,2,3,4,5,6\}$, I'll not include the $0$.

Then we'll get:

$^6C_{1}+^6C_{2}+^6C_{3}+^6C_{4}+^6C_{5}+^6C_{6}=2^6-1$

The minus $1$ is because the $^6C_{0}$ is not inclued. I thing is that.

Explain:

Each combination refers to each set. A set of $1$ element, a set of $2$ elements...and so on.The $^6C_{0}$ it's not valid for this particular problem, because the problem don't allow the empty set. For any line of the Pascal's triangle the sum of all elements of this line is given by $2^n$, where $n$ is the number's line.So if the $^6C_{0}$ is not inclued the solution is given by $2^n-1$, where in this case $n=6$.