How Many 2 Character Domain Names are there?

[Nexus]

I'm a bit rusty on my combinatorics. How many possible "distinct" 2-character alpha-numeric domain names are there (for .com for instance)? I'm still developing DomainFolio, and I used it to run a registrar distribution on all the 2-characters I'd counted, but I'm more than a bit fuzzy on exactly how many there are. I counted 1296. Is that right? Anyone know the formula for getting the appropriate number (not the combinations themselves, but the total number of possible *distinct* combinations.)

Here's the registrar distribution I got:
---------------------------------
Total Domains: 1296
---------------------------------
[617] Network Solutions
[142] Tucows
[115] eNom
[93] Register.com
[60] Bulk Register
[43] Go Daddy Software
[27] Internet Names Worldwide
[24] DotRegistrar.com
[18] DirectNIC
[14] Domain Discover
[14] Dotster
[14] Alldomains.com
[12] Moniker.com
[7] Internetters
[6] Domain Bank
[6] GKG.net
[6] CoreNic
[5] Ascio
[4] GANDI
[4] Dotforce.com
[4] Markmonitor
[3] Wild West Domains
[3] Parava.net
[3] Joker.com
[3] NetNames
[3] Domain The Net
[3] Directi.com
[2] Total Registrations
[2] AITDomains.com
[2] Schlund & Partners
[2] Domain People
[2] OnlineNIC
[2] Paycenter
[2] America Online
[2] Cronon
[2] Domain Zoo
[2] Signature Domains
[1] NamesDirect
[1] 000Domains
[1] NameScout
[1] Fabulous.com
[1] pairNIC
[1] Alice's Registry
[1] Names4Ever
[1] StarGate
[1] NameEngine
[1] Active ISP
[1] Easydns
[1] 007 Names
[1] SRSplus
[1] Domain Registry
[1] BookMyName
[1] 4Domains.com
[1] DomainInfo.com
[1] The Registry at Info Avenue
[1] OVH
[1] NameBay
[1] Netpia
[1] Arsys Internet
[1] NameSecure
---------------------------------

~ Nexus

 

[Ed30]

26x10 + 10x26 = 520 per extension.

 

[Nexus]

Originally posted by Ed30
26x10 + 10x26 = 520 per extension.

Is that calculation right? I can see that logic breaking down after a bit...

Take this: Group 1: "A,B,C" and Group 2: "1,2"

3x2 + 2x3 = 12 combinations?

1. A1
2. A2
3. B1
4. B2
5. C1
6. C2
7. 1A
8. 2A
9. 1B
10. 2B
11. 1C
12. 2C
13. 11
14. 12
>15. Etc, etc...

~ Nexus

 

[Ed30]

It is correct for two character combinations. I have excluded two number and two letter combinations as you didn't ask for those. The combinations for two numbers are:

10x10 = 100

The combinations for two letters are:

26x26 = 676

The combinattions for them all are:

520+100+676 = 1296

 

[dloft]

36x36=1296 a-z,0-9

 

[Nexus]

I was wondering about the two-character combinations, which are all of them. If its 1296, I think that matches the number I got above so I guessing that's it!

Someone else gave me this equation:
Group 1: 26 characters
Group 2: 10 characters
(26 + 10) x (26 + 10) = 1296

Thanks, Ed. If anyone else has anything to add, it'd be appreciated. I still get the feeling I'm blurry on some basic concepts of combinatorical mathematics that aren't really being demonstrated well in this particular case. I thought I remembered something about cubing or the exclamation point.

Is it "take the total amount of characters, and put it to the power of two?" That seems to work too. 36 characters. 36 to the power of 2 = 1296. Not sure how "tight" that notion is (possible exceptions slipping through like duplicate interations or something...)

~ Nexus

 

[Ed30]

All the equations you have been given are correct - they all point to the same answer anyway lol. Good luck with DomainFolio. Let us know when it's finished.

 

[Nexus]

Thanks again, Ed. I just run into these circumstances where I need total combinations, and I keep forgetting the basic methematic principles for always calculating them correctly the first time. Tomorrow I'll probably be re-checking total available 3-letters, or maybe 4 letters. Ug. Anyway, I appreciate the help.

DomainFolio is coming along well. I'm alpha testing with some people now. Turned an important corner this weekend, and its almost ready for beta.

-----------------
EDIT:
I was thinking about it some more. Is the rule summed up best by saying "total number of unique characters SQUARED"?

Is the total number of 3 characters domains: "total number of unique characters CUBED"? (36*36*36=46656) Etc, etc? Basically take the total number of unique characters and put it to the POWER of the amount of characters in the result? 4 Character domains = (36*36*36*36=1,679,616), total amount of 4 LETTER domains = (26*26*26*26=456,976). Total amount of 3 LETTER domains = (26*26*26=17,576). Total amount of 2 LETTER domains = (26*26=676).

Is this correct, or have I fallen off the deep end? If its true, I think I've found the "brain-dead" basic formula I was looking for...
-----------------

~ Nexus

 

[Ed30]

Originally posted by Nexus


Is the total number of 3 characters domains: "total number of unique characters CUBED"? (36*36*36=46656) Etc, etc? Basically take the total number of unique characters and put it to the POWER of the amount of characters in the result? 4 Character domains = (36*36*36*36=1,679,616), total amount of 4 LETTER domains = (26*26*26*26=456,976). Total amount of 3 LETTER domains = (26*26*26=17,576). Total amount of 2 LETTER domains = (26*26=676).

All correct.

 

[BobBob is verified member.]

Originally posted by Nexus
-----------------
EDIT:
I was thinking about it some more. Is the rule summed up best by saying "total number of unique characters SQUARED"?

Is the total number of 3 characters domains: "total number of unique characters CUBED"? (36*36*36=46656) Etc, etc? Basically take the total number of unique characters and put it to the POWER of the amount of characters in the result? 4 Character domains = (36*36*36*36=1,679,616), total amount of 4 LETTER domains = (26*26*26*26=456,976). Total amount of 3 LETTER domains = (26*26*26=17,576). Total amount of 2 LETTER domains = (26*26=676).

Is this correct, or have I fallen off the deep end? If its true, I think I've found the "brain-dead" basic formula I was looking for...
-----------------

~ Nexus

You are on the right track.

However for 3 character names, remember, you can have dashes. Dashes cannot be the first or last character of a domain name, only in the middle.

3 Char names:
36 * 37 * 36 = 47952

4 char names
36 * 37 * 37 * 36 = 1,774,224

Of course if you want the total number of LETTER combinations, you have:

3 LETTER:
26 * 26 * 26 = 17,576

and so on.

-Bob

 

[Nexus]

Thanks, Bob. Because of all the rules involved with dashes and domain names (not to mention appeal), I was just considering alpha-numeric names. Very good point.

~ Nexus

 

[darrenl]

Do you think all 3 letters are registered?

 

[ToastyX]

Originally posted by Darren06
Do you think all 3 letters are registered?

I have checked all 3-letter com, net, and org domains many times. They are all registered, and if I find any available (which only happens under very unusual circumstances these days), I register them.

 

[adoninet]

36^2=1296

 

[Brands&Jingles]

I would certain.ly recommend LL and LLL, especial.ly the jingling ones, those that represent great value, like LLL.me:

More information on http://Dot-Me.Of-cour.se/2009/04/12/3l-me/