Combinatorics: Find the subset I am on the combinatorics unit in my book, and there is a question that I don't even know how to start solving. Can anyone point me in the right direction? It is:

Becky likes to watch DVDs each evening. How many DVDs must she have, if she is able to watch every evening for $24$ consecutive evenings during her winter break? 
a) A different subset of DVDs?
  b) A different subset of three DVDs?

I am not sure what the question means by subset of DVDs and is Becky watching a new DVD every evening or the same DVD? I'm pretty lost.
The answer in the book is a) at least 5 and b) at least 7
 A: Let $S$ is the set of her DVDs and $D$ is its subset, $D \subset S$.
In question a) there is no limit for number of elements of $D$.
In question b) the set $D$ must have exactly 3 elements.
In both cases the subsets $D_1, D_2, \dots, D_{24}$ must be different.
For example, if she had $5$ DVSs, then $S = \{1, 2, 3, 4, 5\}$, and $D_i$ may be:
\begin{aligned}
D_1 &= \{1\} \\
D_2 &= \{1, 2\} \\
D_3 &= \{1, 2, 3, 4, 5\} \\
D_4 &= \{1, 2, 4, 5\} \\
D_5 &= \{1, 2, 3\} \\
D_6 &= \{1, 2, 3, 4\} \\
\end{aligned}
and so on, as no one of subsets is repeated.
So that:


*

*question a) is about the power set of $S$ (the set of all its subsets),

*question b) is about 3-element's combinations.


Let $n$ be the number of (unknown) elements of set $S$. Then:


*

*a) Power set of $S$ has $2^n$ elements (including the empty set).

*b) The number of 3-element's combinations will be $_nC_3 = \frac {n(n-1(n-2)}6$.


In both cases it need be greater of equal 24, so 


*

*for a) the result is $\color{red}5$ (because $2^5 = 32$, while $2^4 = \ only\ 16$),

*for b) the result is $\color{red}7$ (because $_7C_3 = 35$, while $_6C_3 =  \ only\  20$).

