Permutations with sub permutations 
Five friends - Mary, Hannah, John, Tim and Charles - are attending a
  soccer game and are going to sit in a row with 5 consecutive seats.
  Supposing that they sat in a random order, what is the probability of
  John, Charles and Tim sitting together with Hannah between them?

I did:


*

*Total permutations = 5!

*Permutations of John, Charles and Tim with Hannah sitting between =
$3*3*2=18$

*Now I multiply by two, because Mary can sit either on the 1st or last seat $18*2 = 36$

*Probability = $$\frac{36}{5!} = \frac{3}{10}$$


But my book says this is wrong, and the solution should be either:

(a) $\frac{1}{5}$ $$\\$$  (b) $\frac{1}{10}$ $$\\$$  (c) $\frac{1}{20}$ $$\\$$  (d) $\frac{1}{3}$

What did I do wrong?
 A: My interpretation fo the question

what is the probability of John, Charles and Tim sitting together with Hannah between them?

is that the valid groups are these composed of these four people in contact with the condition that Hannah is not on any of the extremes.
Then the group of three men have $3!=6$ different permutations. Each one is multiplied by $2$ because Hannah can sit in second or third place for a total of $6\cdot 2=12$.
Now the last person, Mary, only can sit in one of the extremes, so the total valid permutations are $12\cdot 2=24$, and all possible permutations are $5!=120$.
Then the probability would be $\frac{24}{120}=\frac15=20\%$.

This step

  
*
  
*Permutations of John, Charles and Tim with Hannah sitting between = $3∗3∗2=18$
  

was wrong. Alternatively to what I did above you can accomplish this calculus counting all the permutations between these $4$, that is $4!=24$, and after subtracting the permutations where Hannah is in one extreme, that is $2\cdot 3!=12$ for a total of $24-12=12$.
