Question

As of April 2006, roughly 50 million .com web domain names were registered (e.g., yahoo.com).

a. How many domain names consisting of just two letters in sequence can be formed? How many domain names of length two are there if digits as well as letters are permitted as characters? (Note: A character length of three or more is now mandated.)

b. How many domain names are there consisting of three letters in sequence? How many of this length are there if either letters or digits are permitted? (Note: All are currently taken.)

c. Answer the questions posed in (b) for four-character sequences.

d. As of April 2006, 97,786 of the four-character se - quences using either letters or digits had not yet been claimed. If a four-character name is randomly selected, what is the probability that it is already owned?

Short answer

a) \({\rm{676 and 1296}}\)

b) \({\rm{17576 and 46656}}\)

c) \({\rm{ 456976 and 1679616}}\)

d) \({\rm{P(\{ }}\) four-character name is already owned \({\rm{\} ) = 0}}{\rm{.942}}{\rm{.}}\)

Step-by-step solution

determining the domain names are there consisting of three letters in sequence

The total number of domain names with just two letters (r=2) and characters (n=26) in a sequence may be computed as follows: number of potential outcomes = nr = 262 = 676 a2)

Determining  the domain names of length two are there if digits as well as letters are permitted as characters 

Assume that we can select first element of an ordered pair in \({{\rm{n}}_{\rm{1}}}\)ways, and that for each selected element we can select the second element in \({{\rm{n}}_{\rm{2}}}\)ways, then we have the number of pairs to be \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}\)

\({\rm{26*26 = 676}}\)

Domain names consisting of just two letters in sequence can be formed in ways, because \({{\rm{n}}_{\rm{1}}}{\rm{ = 26}}\)and \({{\rm{n}}_{\rm{2}}}{\rm{ = 26}}\)

If we allow digits as well, we'll get \({\rm{26 + 10 = 36}}\)characters, indicating that we can make a two-character sequence.

\({\rm{36*36 = 1296}}\) ways.

determining the length of there if either letters or digits are permitted

Domain names that consist of three letters in sequence may be made in the same way that domain names consisting of three letters in sequence can be formed in the same way that domain names consisting of three letters in sequence can be formed in the same way that domain names

\({\rm{26*26*26 = 17576}}\)

In the same way as in (a), when we add numbers, we get 36 characters.

\({\rm{36*36*36 = 46656}}\)

in a variety of ways,

To be more specific, for \({\rm{k = 3}}\), we utilize the following:

For k-Tuples, there is a Product Rule.

If we assume that we may pick the first element of an ordered pair in \({{\rm{n}}_{\rm{1}}}\)ways and the second element in \({{\rm{n}}_{\rm{2}}}\) ways for each selected element, then the number of pairings is \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}\)

Similarly, there are \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}{\rm{* \ldots *}}{{\rm{n}}_{\rm{k}}}\) potential \({\rm{k}}\)-tuples given an ordered collection of \({\rm{k}}\) components, where the \({{\rm{k}}^{{\rm{th\;}}}}\) can be selected in \({{\rm{n}}_{\rm{k}}}\) ways.

determining the four-character sequences.

Using the Product Rule for k-Tuplets, the number of ways to build a domain name with four letters in succession is \({\rm{k = }}\) \({\rm{4}}\)

\({\rm{26*26*26*26 = 456976}}\)

ways. In the same way, when we add digits, we get

\({\rm{36*36*36*36 = 1679616}}\)

determining the probability that it is already owned, If a four-character name is randomly selected

The likelihood of a four-character name being held is

\(\begin{aligned}{{\rm{P(\{ four - character name is already owned\} )}}}&{\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{1 - P(\{ four - character name is available\} )}}}\\{}&{\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{1 - }}\frac{{{\rm{97786}}}}{{{\rm{3}}{{\rm{6}}^{\rm{4}}}}}{\rm{ = 0}}{\rm{.942}}}\end{aligned}\)

(1): For each event A, $ \({\rm{P(A) + P}}\left( {{\rm{A'}}} \right){\rm{ = 1}}\), and the complement of the event that the four-character name is already taken is the event that the four-character name is taken;

(2): In the exercise, the number of good outcomes is provided, 97786, and all potential outcomes are stated in (c)

\({\rm{P(A) = }}\frac{{{\rm{\# favorable outcomes inA}}}}{{{\rm{\# outcomes in the sample space}}}}\)