Non Consecutive letters

Going from left to right, let $a$ be the number of letters prior to the "first" of the three letters chosen, $b$ be the number of letters between the first and second letter chosen, $c$ be the number of letters between the second and third letter chosen and $d$ the number of letters after the third letter chosen.

Since $7$ letters are not chosen, we then need to find the number of solutions to the equation $a + b + c + d = 7,$ where $a,b,c,d$ are integers such that $a,d \ge 0$ and $b,c \ge 1.$ (This last condition is to ensure that no two of the letters are consecutive.)

Now let $b' = b - 1$ and $c' = c - 1.$ This then gives us the equation

$a + b' + c' + d = 5$ where $a,b,c,d$ are non-negative integers.

This is now a stars and bars equation, which according to theorem two in the link has

$\dbinom{5 + 4 - 1}{5} = \dbinom{8}{5} = \boxed{56}$ solutions.

First count how many ways are of choosing three different letters so that two or more of the letters are consecutive. There are two cases.

$\text{Case 1: The three letters are consecutive}$ (slide parenthesis)

$8$ ways

$\text{Case 2: Two letters are consecutive}$

$a)$ The $2$ letters are in the edge $\Rightarrow$ can be combined with $7$ letters

$14$ ways

$b)$ The $2$ letters are not in edge $\Rightarrow$ can be combined with $6$ letters (slide parenthesis and green squares)

$42$ ways

Then, do inclusion- exclusion:

$\text{ways of choosing three letters-last cases}$

$\binom{7}{3}-8-14-42=\boxed{56}$ ways