how many subsets of $\{1,2,3,4,5,6,7\}$ has $6$ as largest? Let $A =\{1,2,3,4,5,6,7\}$, The how many subsets have 6 as largest element?
My approach - 
Total subsets are $2^7 = 128$. Then First I have to exclude the $1$ empty subset i.e. $\{\}$, $6$ sets with $1$ element except $\{6\}$, then two elements not containing $6$ as its element(I am not getting a way to count these element) and the $2$ subsets $\{6,7\},\{7,6\}$. Then I am stuck here as I am not getting the way to count these numbers. Any help?
 A: Hint:
Include $6$.  How many ways are there to include zero through five members of $\{1,2,3,4,5\}$?
(Order doesn't matter with sets.)
A: Remember how we determined that the number of subsets was 2 to the power of the number of elements.
We did that because for each element, $a$, either $a$ could be in a subset.  So the total number of sets was the product of all the choices each of which was 2.
This is the same.  Either 1 is in a subset or not.  That 2 choices.  Either 2 is in the subset or not.  That's $2*2$ choice.  Keep it up EXCEPT notice $6$ must be in the subset so that is only $1$ choice and $7$ must not be in the set so that's only one set.  
So the number of sets is $2*2....*2*1*1$ which is what.
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Or
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Don't eliminate the sets one at a time.  Remove ALL the subsets that do not have $6$ in them.  How many do not have $6$.  Remove ALL the subsets that do have $7$.  How many is that?  Then to avoid double counting add back the ones that had $6$ and didn't have $7$.  How many is that?
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Or
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Figure 1/2 of the 128 have 6 and half do not.  That only leaves half of them acceptable.  Then half of those have 7 and half to not.  That leaves only half of those acceptable.
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Or
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No set has $7$.  So all the elements are taken from {$1,2,3,4,5,6$}.
All sets have $6$.  So all elements that aren't $6$ are taken from {$1,2,3,4,5$} and $6$ is always added to it.
There are $2^5$ subsets of {$1,2,3,4,5$} and we are only taking those and sticking a $6$ into them.  There are $2^5$ such sets.
A: Think about it we need $6$ as an element but we can't have $7$ because $7$ is larger than $6$, we can have anything else. 
What we're looking for: How many subsets have $6$ as an element but not $7$? 
We must choose to include $6$, that gives $1$ choice.  We may choose to to include $1$ or not, that gives $2$ choices. We may choose to include $2$ or not, that gives $2$ choices. We may choose to include $3$ or not, that gives $2$ choices. We may choose to include $4$ or not, that gives $2$ choices. We may choose to include $5$ or not, that gives $2$ choices. We must choose not to include $7$, that gives $1$ choice. By the multiplication principle the number subsets with $6$ but not $7$ is:
$$1•2•2•2•2•2•1=2^5$$
