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

ais next tob?

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?