Answered

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


Q2 How many strings of four decimal digits

Ado not cantain the same digit twice?

Bend with even digit ?

Chave exactly three digits that are 9s ?

Verified Answer :

LammettHash

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.

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