# Question on probability of an ordered choice of letters from a larger word

Four letter words will be made from the letters in the word BARNEY. No repetition of letters, and Y is not considered a vowel. What is the probability that two vowels will be side by side in the word?

I'm completely stuck on this question.

• Do you know how many total four-letter words can be made from BARNEY? – G Tony Jacobs Mar 7 '14 at 0:10
• Yeah, 6P4 = 360 – user12949502 Mar 7 '14 at 0:12
• That's a good start; it's what you want for the denominator. For the numerator, figure out how many words you can make from BARNEY with side-by-side vowels. – G Tony Jacobs Mar 7 '14 at 0:13
• I tried considering the two vowels as a single letter, i.e, (5P4 / 6P4) which did not give me the correct answer. – user12949502 Mar 7 '14 at 0:14
• Think of building words with the vowels already in them, like in Rebecca's answer below. Thinking of the vowels as a single letter isn't a bad idea, but it's a letter you're forced to use, so you don't want it to be part of the pool you're choosing from. – G Tony Jacobs Mar 7 '14 at 0:18

## 1 Answer

Hint: There are two vowels A and E. So 4-letter words with them side by side have one of the $6$ follows forms:

AE--
-AE-
--AE
EA--
-EA-
--EA


Given any of these forms, how many ways can we add in the consonants? This gives the total number $N$ of 4-letter words with A and E side by side. And the probability will be $N/^6 P_4=N/360$.