Number of subsets without consecutive numbers Consider $S=\{1,2,\ldots,15\}$. Let $X$ denote the number of subsets of $S$ of four elements which contain no consecutive numbers.
The claim is that $X$ equals the coefficient of $x^{14}$ in $\dfrac{x^6}{(1-x)^5}$. How would I prove this?

Call the elements of your subset $a_1$, $a_2$, $a_2$, $a_4$ in order. Your restrictions translate into:
\begin{align}
x_1 &= a_1 \ge 1 \\
x_2 &= a_2 - a_1 \ge 2 \\
x_3 &= a_3 - a_2 \ge 2 \\
x_4 &= a_4 - a_3 \ge 2 \\
x_5 &= 15 - a_4  \ge 0
\end{align}
The definitons amount to $x_1 + x_2 + x_3 + x_4 + x_5 = 15$. Set up generating functions for each $x_i$ variable:
\begin{align}
x_1:                   &\quad z + z^2 + \cdots = \frac{z}{1 - z} \\
x_2 \text{ to } x_4: &\quad z^2 + z^3 + \cdots = \frac{z^2}{1 - z} \\
x_5:                   &\quad 1 + z + \cdots     = \frac{1}{1 - z}
\end{align}
Multiplying all, and remembering we want the term whose exponent is 15:
\begin{align}
[z^{15}] \frac{z^7}{(1 - z)^5}
   &= [z^8] (1 - z)^{-5} \\
   &= \binom{-5}{8} (-1)^8 \\
   &= 495
\end{align}
