In how ways is it possible to borrow up to five books from a friend who has five different books? My friend got 5 different books. He does not mind how many books I can select to borrow each time - even if I borrow none. How many different selections are possible?
I am thinking that this is a "combination" problem since the order of the books does not matter. 
So I can make these selections:
5 books OR 4 books OR 3 books ...etc OR zero books.
So there would be 6 different selections possible - but this was the wrong answer.
Where have I gone wrong?
 A: If your friend has $n$ different books, you have $2^n$ selections, since for each book you can either borrow or not borrow it. For the number of books you borrow, there are indeed $6$ possibilities here, but the correct answer is greater since two selections with the same number of books may have differing books.
A: What matters is which books you borrow, not how many books you borrow.  Borrowing Apostol's Calculus and Munkres' Topology is different from borrowing Herstein's Algebra and Rudin's Principles of Mathematical Analysis.  
There are $\binom{5}{k}$ ways for you to borrow exactly $k$ of the five books.  Since you can borrow from $0$ to $5$ books, the number of different selections of books you could borrow is
$$\sum_{k = 0}^{5} \binom{5}{k} = \binom{5}{0} + \binom{5}{1} + \binom{5}{2} + \binom{5}{3} + \binom{5}{4} + \binom{5}{5} = 32$$
As Parcly Taxel observed, with each selection, you either choose to borrow a particular text or not borrow it.  Since there are two choices for each of the five books, there are $2^5 = 32$ possible selections of books you could make.
The two approaches are related by the Binomial Theorem.
$$(x + y)^n = \sum_{k = 0}^{n} \binom{n}{k}x^{n - k}y^k$$
If we let $n = 5$, $x = 1$, and $y = 1$, we obtain
$$2^5 = (1 + 1)^5 = \sum_{k = 0}^{5} \binom{5}{k}1^{n - k}1^k = \sum_{k = 0}^{5} \binom{5}{k}$$
A: You are in a way right to say I can borrow 1 book but since all 5 books are different which book in particular you borrow itself can make a difference like 1 book. Can be borrowed in 5 different ways. Let us say books are A,B,C,D,E
When you borrow either you will borrow A on will not borrow A hence book A can have 2 possible ways to share similarly all 5 books.
Hence it will be 2^5.
