Tickets

$1^{st}$ Kaplansky's Lemma: The number of ways we can choose $p$ elements out of a set of $n$ elements such that there's no two adjacent ones, is: $f(n,p)={{n-p+1}\choose{p}}$ Therefore: $f(35,4)={{32}\choose{4}} \Rightarrow \alpha=32, \beta=4.$

Let the 4 ticket numbers be $a, b, c, d$ . Now let there be $v$ ticket numbers less than $a$ , $w$ ticket numbers between $a$ and $b$ , $x$ between $b$ and $c$ , $y$ between $c$ and $d$ , and finally $z$ ticket numbers greater than $d$ .

Now we require that $v + w + x + y + z = 35 - 4 = 31$ , where each of $w, x$ and $y$ must be at least $1$ . Now let $w' = w - 1, x' = x - 1$ and $y' = y - 1$ . Then we have that $v + w' + x' + y' + z = 28$ , where each of the variables can be any value from $0$ to $28$ inclusive. This is a then a "stars and bars" type of problem, and so this last equation has $\binom{32}{4}$ solutions.

Thus the desired quotient is $\frac{32}{4} = \boxed{8}$ .