+0  
 
0
943
2
avatar+28 

English has  letters that can be vowels. This includes Y, which can be either a consonant or a vowel; for this problem, we'll consider Y a vowel.

The other  English letters are always consonants.

How many two-letter "words" can we make from these letters if we are required to use at least one vowel? (We aren't limited to words that have an actual meaning in English. Thus, for this problem, we'll include nonsense "words" like AA, QO, XY, and UZ.)

 Sep 30, 2018
 #1
avatar
+1

There are 5 "true vowels" in the English Language: a, e, i, o, u  plus y for a total of 6.

If repeats are allowed, then you have: 2^6 = 64 "words".

If repeats are not allowed, and position or order is important, then you have: 6P2 =30 "words"

If repeats are not allowed, and position or order is NOT important, then you have: 6C2 =15 "words"

 

Not sure about consonants!!.

 Sep 30, 2018
edited by Guest  Sep 30, 2018
edited by Guest  Sep 30, 2018
 #2
avatar+6248 
+1

This isn't quite correct as it doesn't take into account words with repeated letters.

 

Suppose the first letter is a vowel, there are 26 valid choices for the second letter.  There are 6 vowels.

 

Suppose the first letter is a consonant.  There are 6 valid choices for the second letter.  There are 20 consonants.

 

So the total number of valid words is

 

6 x 26 + 20 x 6 = 276

Rom  Sep 30, 2018

7 Online Users

avatar
avatar
avatar
avatar
avatar
avatar