1. Condition for three numbers to be in A.P.

If $\binom{n}{4}, \binom{n}{5}, \binom{n}{6}$ are in A.P., then

$2\binom{n}{5}=\binom{n}{4}+\binom{n}{6}.$

2. Write the combinations in comparable form

Using the ratio relations:

$\binom{n}{5}=\binom{n}{4}\cdot \frac{n-4}{5}$

and

$\binom{n}{6}=\binom{n}{5}\cdot \frac{n-5}{6}=\binom{n}{4}\cdot \frac{(n-4)(n-5)}{30}.$

Substitute into the A.P. condition:

$2\binom{n}{4}\cdot \frac{n-4}{5}=\binom{n}{4}+\binom{n}{4}\cdot \frac{(n-4)(n-5)}{30}.$

Since $\binom{n}{4}\neq 0$ for valid $n\ge 6$, divide throughout by $\binom{n}{4}$:

$\frac{2(n-4)}{5}=1+\frac{(n-4)(n-5)}{30}.$

Multiply by $30$:

$12(n-4)=30+(n-4)(n-5).$

Expand:

$12n-48=30+n^2-9n+20.$

$12n-48=n^2-9n+50.$

Bring all terms to one side:

$n^2-21n+98=0.$

Factorize:

$n^2-21n+98=(n-14)(n-7)=0.$

So,

$n=14 \text{ or } 7.$

3. Check with the given options

Options are:

• A: $11$
• B: $12$
• C: $9$
• D: $14$

Among these, only $14$ is present.

Hence the correct option is

$\boxed{14}.$

4. Comparison with stored answer

Stored correct answer is D, i.e. $14$.

This matches our result.