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

#1**+1 **

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.

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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,

Gavin

GYanggg
May 18, 2018

#2**0 **

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

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

Guest May 18, 2018

#4**+1 **

**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.

GA

GingerAle
May 18, 2018