By definition, $ r! \equal{} r(r \minus{} 1) \cdots 1$ and $ \binom{j}{k} \equal{} \frac {j!}{k!(j \minus{} k)!}$, where $ r,j,k$ are positive integers and $ k < j$. If $ \binom{n}{1}, \binom{n}{2}, \binom{n}{3}$ form an arithmetic progression with $ n > 3$, then $ n$ equals 

$ \textbf{(A)}\ 5\qquad
\textbf{(B)}\ 7\qquad
\textbf{(C)}\ 9\qquad
\textbf{(D)}\ 11\qquad
\textbf{(E)}\ 12$
1. We start by expressing the binomial coefficients $\binom{n}{1}$, $\binom{n}{2}$, and $\binom{n}{3}$ in terms of $n$:
   \[
   \binom{n}{1} = n
   \]
   \[
   \binom{n}{2} = \frac{n(n-1)}{2}
   \]
   \[
   \binom{n}{3} = \frac{n(n-1)(n-2)}{6}
   \]

2. Since these coefficients form an arithmetic progression, the difference between consecutive terms must be constant. Therefore, we set up the following equation:
   \[
   \binom{n}{2} - \binom{n}{1} = \binom{n}{3} - \binom{n}{2}
   \]

3. Substitute the expressions for the binomial coefficients:
   \[
   \frac{n(n-1)}{2} - n = \frac{n(n-1)(n-2)}{6} - \frac{n(n-1)}{2}
   \]

4. Simplify the left-hand side:
   \[
   \frac{n(n-1)}{2} - n = \frac{n(n-1)}{2} - \frac{2n}{2} = \frac{n(n-1) - 2n}{2} = \frac{n^2 - n - 2n}{2} = \frac{n^2 - 3n}{2}
   \]

5. Simplify the right-hand side:
   \[
   \frac{n(n-1)(n-2)}{6} - \frac{n(n-1)}{2} = \frac{n(n-1)(n-2) - 3n(n-1)}{6} = \frac{n(n-1)(n-2) - 3n(n-1)}{6}
   \]
   Factor out $n(n-1)$:
   \[
   \frac{n(n-1)(n-2 - 3)}{6} = \frac{n(n-1)(n-5)}{6}
   \]

6. Equate the simplified expressions from both sides:
   \[
   \frac{n^2 - 3n}{2} = \frac{n(n-1)(n-5)}{6}
   \]

7. Clear the fractions by multiplying both sides by 6:
   \[
   3(n^2 - 3n) = n(n-1)(n-5)
   \]
   \[
   3n^2 - 9n = n^3 - 6n^2 + 5n
   \]

8. Rearrange the equation to set it to zero:
   \[
   0 = n^3 - 6n^2 + 5n - 3n^2 + 9n
   \]
   \[
   0 = n^3 - 9n^2 + 14n
   \]

9. Factor out $n$:
   \[
   0 = n(n^2 - 9n + 14)
   \]

10. Solve the quadratic equation $n^2 - 9n + 14 = 0$ using the quadratic formula:
    \[
    n = \frac{9 \pm \sqrt{81 - 56}}{2} = \frac{9 \pm \sqrt{25}}{2} = \frac{9 \pm 5}{2}
    \]
    \[
    n = \frac{14}{2} = 7 \quad \text{or} \quad n = \frac{4}{2} = 2
    \]

11. Since $n > 3$, we discard $n = 2$ and accept $n = 7$.

The final answer is $\boxed{7}$