How many 2-letter "words" consist of two different letters arranged in alphabetical order? (Any two letters together is considered a "word." For example, one such word is DQ.)

rohitaop  May 18, 2018

Hey rohitaop!


Here is my solution:


We can consider the cases:


If A is the first letter, we have B-Z possibilities, or 25 letters. 


If B is the first letter, we have C-Z possibilities, or 24 letters. 




If Y is the first letter, we have Z possibilities, or 1 letter. 


We have the sum:


1 + 2 + 3 + 4 ... + 24 + 25, we use the arithmetic sum formula where the difference is 1. 


25 * (25+1) / 2 = 325.


I hope this helped,



GYanggg  May 18, 2018

What is so "special" about these combinations? They are simple binomials:

26C2 = 325, in alphabetical order: AB, AC, AD.........etc.

Guest May 18, 2018

No! Mr. BB., they are not simple binomials in alphabetical order. Gobble! Gobble!

GingerAle  May 18, 2018

This is not the correct solution for this question.

For combinations, a set such as AB = BA, and the two sets are the same and counted only once. If this were the case, then 26C2 would be the correct solution. \(26C2 = \dfrac {26!}{2!(26-2)!} = 325\)


In this question, the order of the sets is unique and counting the permutations of these sets gives the correct solution.

\(26P2 = \frac {26!}{(26n-2)!} = 650\)

Note that the formulas differ only by the division of the set size factorial (2!).


Here’s another method for solving.

Counting all combinations gives \(26^2 = 676\) sets of two letters including AA , BB, ect.

The double letters occurs once for each of the 26 letters. Subtracting the 26 duplicates from 676 leaves 650 sets, and this the number of sets without a repeated letter.


Note also that alphabetizing does not make a difference in the set counts. 




GingerAle  May 18, 2018
edited by GingerAle  May 18, 2018

How many 2-letter "words" consist of two different letters arranged in alphabetical order?



the guest that you were so eager to criticize was right.

Guest May 21, 2018

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