A small library has exactly 3 books: A, B,&C.In how many ways can you check out none,some or all of the books?

This is a math problem given to me. I've come up with a couple different answers:( Can someone help, please?

A small library has exactly 3 books: A, B,%26amp;C.In how many ways can you check out none,some or all of the books?
For book A you have two options (to check it out or not check it out).





For book B you have two options (to check it out or to not check it out).





For book C you have two options (to check it out or to not check it out).





So to find out your total options you multiply your options for each book (this is known as the multiplication principle of combinatorics) and you get





2 * 2 * 2 = 2^3 = 8.





You could also of course list out all the possibilities, but it's easy to forget one (or more).





1. none or { }





2. {A}





3. {A, B}





4. {A, B, C}





5. {A, C}





6. {B}





7. {B, C}





8. {C}





In general if you have the option of checking out n books there are a total of





2^n





ways of checking out none, some, or all of the books.





Mathematically, this is equivalent to saying the number of subsets of a set of n elements is 2^n.
Reply:If all the three are different books :


none can be done in C(3, 0) = 1 way only that is checking out none.


one in C(3, !) = 3 ways i.e. either A or B or C.


two in C(3, 2) = 3 way i.e. either A,B or A,C or B,C.


all in C(3, 3) = 1 way only i.e. A,B,C.
Reply:1 way to check out none


3 ways to check out 1 = a, b , %26amp; c


3 ways to check out 2: ab, ac, bc


1 way to check out 3 : abc
Reply:1. none


2. A


3. B


4. C


5. A and B


6. A and C


7. B and C


8. A and B and C





In data management it is phrased as follows:


"x choose y"





3C0 = 1


3C1 = 3


3C2 = 3


3C3 = 1





Add these up and you get 8.


You can also get more answers for these "choose" questions by typing in "3 choose 1" (for example) in a google search