How many different subsets of a $10$-element set are there where the subsets have at most $8$ elements? How many different subsets of a $10$-element set are there where the subsets have at most $8$ elements?
How do I do this problem?
 A: The number of $9$-element subsets is $10\choose9$, and the number of $10$-element subsets is $10\choose10$.
So the required ans is $2^{10}$-$10\choose9$-$10\choose10$$=1024-10-1=1013$
A: If order matters, what you are referring to is known as a permutation.
https://en.wikipedia.org/wiki/Permutation
If order does not matter, it is a combination. 
https://en.wikipedia.org/wiki/Combination
Through google, you can find numerous explanations of them. 
A: The power set has $2^{10}$ elements. Then, the number of subsets with exactly $10$ elements would the set itself, in other words $\begin{pmatrix} 10 \\ 10 \end{pmatrix}$ subsets. Then, the number of subsets with exactly $9$ elements would be all of the elements minus one arbitrary element, since there are $10$ elements we have $10$ subsets with this property, in other words $\begin{pmatrix} 10 \\ 9 \end{pmatrix}$ subsets. Hence, the amount of subsets with at most $8$ subsets would be:
$$2^{10} - 10 - 1 = 1013.$$
A: If a set has $n$ elements and you need to count the number of subsets with exactly $m$ elements we reason ... there are $n$ possible elements we can choose from the main set to put in the subset; then from the remaining $n-1$ we can $n-1$ possible to also put in the set;  from the remaining $n-2$ we can choose any to also put in the set.  So in the end there were $n*(n-1)*(n-2)*.......*(n-m+1)$ ways to pick what elements to put in the subset. But VERY IMPORTANT those choices all place the elements in a particular order which is not relevent-- if we put apples and oranges into a shopping bag it doesn't matter if we put the apples in first or the oranges in first.  So for every subset with $m$ elements there are $m!$ ways to arrange them ($m$ choices for the first elemet, $m-1$ for the second, etc.) becauuse $n*..... *(n-m+1)$ has every possible order and order doesn't matter, we have $m!$ too many.  So the number of possible subsets with $m$ elements are:
$\frac {n*..... *(n-m+1)}{m!} =$
$\frac {n*(n-1)*....*(n-m+1)*(n-m)...*2*1}{(n-m)*......*2*1)*m!}=$
$ \frac {n!}{(n-m)!m!} = {n \choose m}$
The notation ${n \choose m}$ is defined as $ \frac {n!}{(n-m)!m!}$ is called "$n$ choose $m$" and it literally means the number of ways to choose $m$ elements for set of $n$ elements.  So by definition then number of subsets with $m$ elements from a master set of $n$ elements is ${n \choose m} =\frac {n!}{(n-m)!m!}$.
So the number of subsets with at most $8$ elements are: The number of subsets with 0 elements + the number of subsets with 1 element + the number of subsets with 2 elements + .... + the number of subsets with 8 elements or:
${10 \choose 0} + {10 \choose 1} + {10 \choose 2} + ...... + {10 \choose 8}$.
But that is a lot to calculate.  There is probably an easier way.
Let $T = \text{the total number of all subsets}$ 
Then the number of subsets with at most $8$ elements are $ T - \text{the number of sets with 9 or 10 elements} = T - {10 \choose 9} - {10 \choose 10}$.
${10 \choose 9} = \frac {10!}{9!*1!} = \frac {10*9*8*.....*1}{9*8*....*1} =10$.  That is not surprising as the number as to get an subset with $9$ elements we  have to remove an element, and there are $10$ elements to choose from.
${10 \choose 10} = \frac {10!}{10!0!} = \frac {10!}{10!*1} = 1$.  This is absolutely not surprising as the only subset with 10 elements has to be the set itself.
So the number of subsets with at most $8$ elements is $T - 11$.
But what is $T$?
If a set has $n$ elements, for each element you have a choose: either you put the element into a subset, or you don't put it into a subset.  So there are $2^n$ possible subsets you can make.
So $T = 2^{10} = 1024$ and $11 $ of them have $9$ or more elements.  So the number of sets with $8$ or fewer are $1024 -11 = 1013$.
