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Here's a question from a book on probability I'm working through:

If the $26$ letters of the alphabet are written down in a ring so that no two vowels come together, what is the chance that a is next to b?

Here's what I did. Let's fix a. Since b can be immediately to the left or right of a, there's $2$ choices for b. Without loss of generality let's say we have ab, and so we have $4$ vowels remaining and $20$ consonants remaining. With the condition that no $2$ vowels come together: We want to find the number of possible places where we can place a vowel, which is in between consonants. With the $20$ consonants, there's $19$ possible "gaps" between, plus the $2$ on the end, for a total of $21$. However, because our b already occupies one of them, we have to subtract $1$, getting us $20$. So out of these $20$ places we're choosing $4$, so there's $\binom{20}{4}$ ways to place our $4$ vowels among the $20$ consonants subject to the condition no $2$ vowels come together. There's $4!$ ways to order the remaining vowels and $20!$ ways to order the remaining consonants. So our numerator is$${{2 \binom{20}{4} 20!4!}}$$Now let's calculate the denominator. Fix a again. This time we have $21$ consonants and $4$ vowels remaining, but only the $20$ possible "gaps" between the consonants to places our $4$ vowels (there's no $2$ at the end this time around), so our numerator is$$\binom{20}{4}21!4!$$Therefore the chance that a is next to b is$${2\over{21}}$$However, the answer in the back of my book (which is known to be wrong in many places in the answers in the back section) is ${1\over{10}}$. So who's correct? And if I'm wrong, where did I specifically go wrong?

RobPratt
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Emperor Concerto
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  • I _think_ you might have a fencepost issue. If you imagine the 20 consonants in a straight line, then the gap before the first is the same as the gap after the last, since they loop around and connect. – Alan Aug 14 '21 at 22:08
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    Note that you would get $\frac 1{10}$ if you treated "y" as a vowel. So if the rule for vowels is "a,e,i,o,u, and sometimes y" then sometimes you are right and sometimes the book is right (: – WW1 Aug 14 '21 at 22:21
  • @Emperor Concerto Can you please write the name of that book ? – user955791 Aug 15 '21 at 07:49

1 Answers1

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A simpler reasoning: We have $26-5=21$ consonants. We know that, in the allowed configurations, each vowel (in particular, $a$) will have two consonants on each side. By symmetry, any pair of consonants is equiprobable. There are $\binom{21}{ 2}=210$ such (unordered) pairs. Among them, there are $20$ pairs that include $b$. Then the probability of our event (letter $a$ is next to $b$) is

$$ p=\frac{20}{210}=\frac{2}{21}=0.95238\cdots$$

which agrees with your answer.

leonbloy
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