A set of [tex]n[/tex] elements has [tex]2^n[/tex] possible subsets, where two classes of those sets are the empty set (1) and all the possible singleton sets ([tex]n[/tex]). So a set of [tex]n[/tex] elements has [tex]2^n-1-n[/tex] possible subsets with more than one elements. For Q1 take [tex]n=100[/tex].
Q2a. Assuming not containing the same digits twice also includes not numbers with three or four of the same digit, and assuming digits are chosen from the usual 0-9, there are
[tex]4!\dbinom{10}4=\dfrac{4!10!}{4!(10-4)!}=10\times9\times8\times7=5040[/tex]
possible strings.
Q2b. The first three digits can be chosen freely from 0-9, while the last digit has to be one of 0, 2, 4, 6, or 8. This means you have
[tex]10^3\times5=5000[/tex]
possible strings
Q2c. Any such string will take the form [tex]999X[/tex], [tex]99X9[/tex], [tex]9X99[/tex], or [tex]X999[/tex], where [tex]X[/tex] has 9 possible choices (0-9 excluding 9, since we want exactly three 9s in any such string). So there are
[tex]4\times9=36[/tex]
possible strings.
