Question

Q1 How many subsets of a set with 100 elements have morethan one element ?

Answer 1

A set of $n$ elements has $2^n$ possible subsets, where two classes of those sets are the empty set (1) and all the possible singleton sets ($n$). So a set of $n$ elements has $2^n-1-n$ possible subsets with more than one elements. For Q1 take $n=100$.

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

$4!\dbinom{10}4=\dfrac{4!10!}{4!(10-4)!}=10\times9\times8\times7=5040$

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

$10^3\times5=5000$

possible strings

Q2c. Any such string will take the form $999X$, $99X9$, $9X99$, or $X999$, where $X$ has 9 possible choices (0-9 excluding 9, since we want exactly three 9s in any such string). So there are

$4\times9=36$

possible strings.